HC Verma ii Solutions for Class 12 Science Physics Chapter 24 Kinetic Theory Of Gases are provided here with simple step-by-step explanations. These solutions for Kinetic Theory Of Gases are extremely popular among class 12 Science students for Physics Kinetic Theory Of Gases Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma ii Book of class 12 Science Physics Chapter 24 are provided here for you for free. You will also love the ad-free experience on Meritnation’s HC Verma ii Solutions. All HC Verma ii Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

Page No 32:

Question 1:

Answer:

No, kinetic energy of the molecules does not increase. This is because velocity of the molecules does not increase with respect to the walls of the gas cylinder, when the cylinder is kept in a vehicle moving with a uniform motion. However, if the vehicle is accelerated or decelerated, then there will be a change in the gas's kinetic energy and there will be a rise in the temperature.  

Page No 32:

Question 2:

No, kinetic energy of the molecules does not increase. This is because velocity of the molecules does not increase with respect to the walls of the gas cylinder, when the cylinder is kept in a vehicle moving with a uniform motion. However, if the vehicle is accelerated or decelerated, then there will be a change in the gas's kinetic energy and there will be a rise in the temperature.  

Answer:

Inside a cooking gas cylinder, the gas is kept in the liquid state using high pressure. Boiling point of a liquid depends on the pressure above its surface. Higher the pressure above the liquid, higher will be its boiling point.

When the gas oven is switched on, the vapour pressure inside the cylinder decreases. To compensate this fall in pressure, more liquid undergoes phase transition (vapourisation) to build up the earlier pressure. In this way, more and more gas evaporates from the liquified state at constant pressure.    

Page No 32:

Question 3:

Inside a cooking gas cylinder, the gas is kept in the liquid state using high pressure. Boiling point of a liquid depends on the pressure above its surface. Higher the pressure above the liquid, higher will be its boiling point.

When the gas oven is switched on, the vapour pressure inside the cylinder decreases. To compensate this fall in pressure, more liquid undergoes phase transition (vapourisation) to build up the earlier pressure. In this way, more and more gas evaporates from the liquified state at constant pressure.    

Answer:

No, the gas won't obey ideal gas equation due to the following reasons:

1. In a cooking gas cylinder, the gas is kept at high pressure and at room temperature. Real gases behave ideally only at low pressure and high temperature.
2. Cooking gas is kept in liquid state inside the cylinder becaue liquid state does not obey the ideal gas equation. 

Page No 32:

Question 4:

No, the gas won't obey ideal gas equation due to the following reasons:

1. In a cooking gas cylinder, the gas is kept at high pressure and at room temperature. Real gases behave ideally only at low pressure and high temperature.
2. Cooking gas is kept in liquid state inside the cylinder becaue liquid state does not obey the ideal gas equation. 

Answer:

(a) Temperature is defined as the average kinetic energy of he partciles. In a vacuum, devoid of any electromagnetic fields and molecules or entities, the temperature cannot be defined as there are no molecules or atoms or entities. 

(b) No, we cannot define temperature of a single molecule. Since temperature is defined as the average kinetic energy of the particles, it is defined only statistically for a large collection of molecules. 

Page No 32:

Question 5:

(a) Temperature is defined as the average kinetic energy of he partciles. In a vacuum, devoid of any electromagnetic fields and molecules or entities, the temperature cannot be defined as there are no molecules or atoms or entities. 

(b) No, we cannot define temperature of a single molecule. Since temperature is defined as the average kinetic energy of the particles, it is defined only statistically for a large collection of molecules. 

Answer:

Yes, at equilibrium all the molecules in a sample of gas have the same temperature. This is because temperature is defined as the average kinetic energy for all the molecules in a system. Since all the molecules have the same average, temperature will be the same for all the molecules. 

Page No 32:

Question 6:

Yes, at equilibrium all the molecules in a sample of gas have the same temperature. This is because temperature is defined as the average kinetic energy for all the molecules in a system. Since all the molecules have the same average, temperature will be the same for all the molecules. 

Answer:

Yes, according to the postulates of kinetic theory, a gas of neutrons will be a better ideal gas than hydrogen. The reasons are given below:

1. As per the Kinetic theory, neutrons do not interact with each othe. Molecules of an ideal gas should also not interact with each other. On the other hand, hydrogen molecules interact with each other owing to the presence of charges in them.

2. Neutrons are smaller than hydrogen. This fulfils another kinetic theory postulate that gas molecules should be points and should have negligible size. 

Page No 32:

Question 7:

Yes, according to the postulates of kinetic theory, a gas of neutrons will be a better ideal gas than hydrogen. The reasons are given below:

1. As per the Kinetic theory, neutrons do not interact with each othe. Molecules of an ideal gas should also not interact with each other. On the other hand, hydrogen molecules interact with each other owing to the presence of charges in them.

2. Neutrons are smaller than hydrogen. This fulfils another kinetic theory postulate that gas molecules should be points and should have negligible size. 

Answer:

No, the pressure on gas won't increase because of this. The pressure will not be transferred to the gas, but to the container and to the ground.  

Page No 32:

Question 8:

No, the pressure on gas won't increase because of this. The pressure will not be transferred to the gas, but to the container and to the ground.  

Answer:

Since the pressure would be zero, the molecules would not collide with the walls and would not transfer momentum to the walls. This is because pressure of a gas is formed due to the molecule's collision with the walls of the container.   

Page No 32:

Question 9:

Since the pressure would be zero, the molecules would not collide with the walls and would not transfer momentum to the walls. This is because pressure of a gas is formed due to the molecule's collision with the walls of the container.   

Answer:

Two postulates of kinetic theory will not be valid in this case. These are given below:

1. All gases are made up of molecules moving randomly in all directions
2. When a gas is left for a sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction and time. 

Page No 32:

Question 10:

Two postulates of kinetic theory will not be valid in this case. These are given below:

1. All gases are made up of molecules moving randomly in all directions
2. When a gas is left for a sufficient time, it comes to a steady state. The density and the distribution of molecules with different velocities are independent of position, direction and time. 

Answer:

If the gas is ideal, there will be no temperature change. Moreover, Charles's law relates volume with temperature not pressure with temperature, so the cause behind the phenomena cannot be explained by Charles's law.

Page No 32:

Question 11:

If the gas is ideal, there will be no temperature change. Moreover, Charles's law relates volume with temperature not pressure with temperature, so the cause behind the phenomena cannot be explained by Charles's law.

Answer:

In a pressure cooker, the vapour pressure over the water surface is more than the atmospheric pressure. This means boiling point of the water will be higher in the pressure cooker than in the open. This will let the cereals and food to be cooked in higher temperature than at 1000C,. Thus, cooking process gets faster.

Page No 32:

Question 12:

In a pressure cooker, the vapour pressure over the water surface is more than the atmospheric pressure. This means boiling point of the water will be higher in the pressure cooker than in the open. This will let the cereals and food to be cooked in higher temperature than at 1000C,. Thus, cooking process gets faster.

Answer:

If the molecules are not allowed to collide with each other, they will have long mean free paths and hence, evaporation will be faster. In vacuum, the external pressure will be very low. So, the liquid will boil and evaporate at very low temperature.   

Page No 32:

Question 13:

If the molecules are not allowed to collide with each other, they will have long mean free paths and hence, evaporation will be faster. In vacuum, the external pressure will be very low. So, the liquid will boil and evaporate at very low temperature.   

Answer:

Yes, it is possible to boil water at 300C by reducing the external pressure. A liquid boils when its vapour pressure equals external pressure. By lowering the external pressure, it is possible to boil the liquid at low temperatures.

No, the flask containing water boiling at 300C will not be hot.

Page No 32:

Question 14:

Yes, it is possible to boil water at 300C by reducing the external pressure. A liquid boils when its vapour pressure equals external pressure. By lowering the external pressure, it is possible to boil the liquid at low temperatures.

No, the flask containing water boiling at 300C will not be hot.

Answer:

After a dip in the river, the water that sticks to our body gets evaporated. We know that evaporation takes place faster for higher temperatures. Thus, the molecules that have the highest kinetic energy leave faster and that is how heat is given away from our body.
As a result of it, temperature of our body falls down due to loss of heat and we feel cold.



Page No 33:

Question 1:

After a dip in the river, the water that sticks to our body gets evaporated. We know that evaporation takes place faster for higher temperatures. Thus, the molecules that have the highest kinetic energy leave faster and that is how heat is given away from our body.
As a result of it, temperature of our body falls down due to loss of heat and we feel cold.

Answer:

(d) Kinetic energy.

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.

Hence, correct answer is (d).

Page No 33:

Question 2:

(d) Kinetic energy.

Temperature is defined as the average kinetic energy of the molecules in a gas sample. Average is same for all the molecules of the sample. So, kinetic energy is the same for all.

Hence, correct answer is (d).

Answer:


At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies. 
At high temperature, molecules move very fast. So, they tend to collide elastically and ​forces of interaction between the molecules minimise. This is the required idea condition.

Thus, (b) is the correct answer.

Page No 33:

Question 3:


At low pressure, the concentration of gas molecules is very low. Hence, the kinetic assumption that the size of the molecules can be neglected compared to the volume of the container applies. 
At high temperature, molecules move very fast. So, they tend to collide elastically and ​forces of interaction between the molecules minimise. This is the required idea condition.

Thus, (b) is the correct answer.

Answer:


According to the kinetic theory, molecules show straight line in motion (translational).  So, the kinetic energy is essentially transitional. 

Thus, (a) is the correct answer.

Page No 33:

Question 4:


According to the kinetic theory, molecules show straight line in motion (translational).  So, the kinetic energy is essentially transitional. 

Thus, (a) is the correct answer.

Answer:


Temperature of a gas is directly proportional to its kinetic energy. Thus, energy of an ideal gas depends only on its temperature. 

Thus, (d) is the correct answer.

Page No 33:

Question 5:


Temperature of a gas is directly proportional to its kinetic energy. Thus, energy of an ideal gas depends only on its temperature. 

Thus, (d) is the correct answer.

Answer:


The rms speed of a gas is given by â€‹3RTMo.
Since hydrogen has the lowest Mo compared to other molecules, it will have the highest rms speed. 
Thus, (a) is the correct answer.

Page No 33:

Question 6:


The rms speed of a gas is given by â€‹3RTMo.
Since hydrogen has the lowest Mo compared to other molecules, it will have the highest rms speed. 
Thus, (a) is the correct answer.

Answer:


The straight line T1 has greater slope than T​2. This means Pρ ratio is greater for T1 than T2. Now, rms velocity of a gas is given by 3Pρ. This means rms velocity of gas with T1 molecules is greater than T2 molecules. Again, gas with higher temperature has higher rms velocity.

So, T1 > T2.

Thus, (a) is the correct answer.

Page No 33:

Question 7:


The straight line T1 has greater slope than T​2. This means Pρ ratio is greater for T1 than T2. Now, rms velocity of a gas is given by 3Pρ. This means rms velocity of gas with T1 molecules is greater than T2 molecules. Again, gas with higher temperature has higher rms velocity.

So, T1 > T2.

Thus, (a) is the correct answer.

Answer:


Root mean squared velocity is given by
vrms=3RTMvrms2=3RTMvrms2 α T

Thus, (c) is the correct answer.

Page No 33:

Question 8:


Root mean squared velocity is given by
vrms=3RTMvrms2=3RTMvrms2 α T

Thus, (c) is the correct answer.

Answer:


Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed.

Thus, (c) is the correct answer.

Page No 33:

Question 9:


Speed is constant and same for a single molecule. Thus, rms speed will be equal to its average speed.

Thus, (c) is the correct answer.

Answer:


(b) 2000 ms−1

Given,
Molecular mass of hydrogen, MH = 2
Molecular mass of oxygen, Mo = 32
RMS speed is given by,

 vrms=3RTM 3RTMO=500
Now,
vOrmsvHrms=3RTMO3RTMHvOrmsvHrms=3RT323RT2vOrmsvHrms=14500vHrms=14vHrms=4×500=2000 ms-1

Hence, the correct answer is (b).

Page No 33:

Question 10:


(b) 2000 ms−1

Given,
Molecular mass of hydrogen, MH = 2
Molecular mass of oxygen, Mo = 32
RMS speed is given by,

 vrms=3RTM 3RTMO=500
Now,
vOrmsvHrms=3RTMO3RTMHvOrmsvHrms=3RT323RT2vOrmsvHrms=14500vHrms=14vHrms=4×500=2000 ms-1

Hence, the correct answer is (b).

Answer:


Let the number of moles in the gas be n.
Applying equation of state, we get

PV=nRTP=nRTV2×105=nRTV                      ...1When half of the gas is removed, number of moles left behind =n2Let the pressure be P'.P'=n2RTVNow,P'=12×2×105=105                   From eq. 1
=100 kPa

Thus, (a) is the correct answer.

Page No 33:

Question 11:


Let the number of moles in the gas be n.
Applying equation of state, we get

PV=nRTP=nRTV2×105=nRTV                      ...1When half of the gas is removed, number of moles left behind =n2Let the pressure be P'.P'=n2RTVNow,P'=12×2×105=105                   From eq. 1
=100 kPa

Thus, (a) is the correct answer.

Answer:


Given: v=3RT32Let the new rms speed be v'.Molecule dissociate, M=16v'=3R2T16=3R4T32=23RT32=2v

Thus, (c) is the correct answer.

Page No 33:

Question 12:


Given: v=3RT32Let the new rms speed be v'.Molecule dissociate, M=16v'=3R2T16=3R4T32=23RT32=2v

Thus, (c) is the correct answer.

Answer:


Here,PV=nRT                      ...1Also,k=RNR=kN                      ...2Now,PV=nkNT                        From eq. 1 and eq. 2nN=PVkTnN = Number of moleculesPVkT=Number of molecules

Thus, (d) is the correct answer.

Page No 33:

Question 13:


Here,PV=nRT                      ...1Also,k=RNR=kN                      ...2Now,PV=nkNT                        From eq. 1 and eq. 2nN=PVkTnN = Number of moleculesPVkT=Number of molecules

Thus, (d) is the correct answer.

Answer:


According to the graph, P is directly proportional to T.

Applying the equation of state, we get

PV = nRT
P=nRVTGiven: P α TThis means nRV is a constant. So, V is also a constant.

Constant V implies the process is isochoric.

Thus, (c) is the correct answer.

Page No 33:

Question 14:


According to the graph, P is directly proportional to T.

Applying the equation of state, we get

PV = nRT
P=nRVTGiven: P α TThis means nRV is a constant. So, V is also a constant.

Constant V implies the process is isochoric.

Thus, (c) is the correct answer.

Answer:


As the liquid is decreasing, the liquid is vapourised. We know that vapourisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.  

Thus, (b) is the correct answer.

Page No 33:

Question 15:


As the liquid is decreasing, the liquid is vapourised. We know that vapourisation cannot occur in saturated air and there cannot be any liquid with no vapour at all. So, the vapour in the remaining part is unsaturated.  

Thus, (b) is the correct answer.

Answer:

Since the amount of liquid is constant, there is no vapourisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vapourisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid. 

Thus, (a) is the correct answer.

Page No 33:

Question 16:

Since the amount of liquid is constant, there is no vapourisation of the liquid inside the bottle. Also, since there cannot be a liquid with no vapours at all and vapourisation cannot take place in the remaining saturated part, the remaining part must be saturated with the vapours of the liquid. 

Thus, (a) is the correct answer.

Answer:


As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Thus, (d) is the correct answer.

Page No 33:

Question 17:


As the vapour is injected, the pressure of the chamber increases. But when the pressure becomes equal to the saturated vapour pressure, it condenses. So, if more vapour is injected beyond the saturated vapour pressure, the vapour will condense and thus the vapour pressure will be constant.

Thus, (d) is the correct answer.

Answer:

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume,  both the vessels will have same pressure.

Thus, (a) is the correct answer.

Page No 33:

Question 1:

The maximum pressure attainable above the water will be saturated vapour pressure at that temperature. Since saturated vapour pressure does not depend upon volume,  both the vessels will have same pressure.

Thus, (a) is the correct answer.

Answer:


According to Kinetic theory, postulates collision between molecules are elastic. This means that kinetic energy after any collision is conserved because while one one gains kinetic energy, another loses it. Both options, (c) and (d) consider the conservation of kinetic energy in the collision. 

Thus, (c) and (d) are correct answers.



Page No 34:

Question 2:


According to Kinetic theory, postulates collision between molecules are elastic. This means that kinetic energy after any collision is conserved because while one one gains kinetic energy, another loses it. Both options, (c) and (d) consider the conservation of kinetic energy in the collision. 

Thus, (c) and (d) are correct answers.

Answer:

The average speed of molecules is given by 8kTπm. We observe that greater the mass, lesser is the average speed of the molecule. Since an oxygen molecule is heavier than a hydrogen molecule, the oxygen molecule will hit the wall with smaller average speed.

Thus, (b) is the correct answer.

Page No 34:

Question 3:

The average speed of molecules is given by 8kTπm. We observe that greater the mass, lesser is the average speed of the molecule. Since an oxygen molecule is heavier than a hydrogen molecule, the oxygen molecule will hit the wall with smaller average speed.

Thus, (b) is the correct answer.

Answer:


The molecules move in all possible directions in an ideal gas at equilibrium. Since momentum is a vector quantity for every direction of motion of the molecules, there exists an opposite direction of motion of the other. Hence, the average momentum is zero for an ideal gas at equilibrium.

Thus, (b) is the correct answer.

Page No 34:

Question 4:


The molecules move in all possible directions in an ideal gas at equilibrium. Since momentum is a vector quantity for every direction of motion of the molecules, there exists an opposite direction of motion of the other. Hence, the average momentum is zero for an ideal gas at equilibrium.

Thus, (b) is the correct answer.

Answer:


Pressure of an ideal gas is given by PV = 13mnu2. We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change. 

Thus, (c) is the correct answer.

Page No 34:

Question 5:


Pressure of an ideal gas is given by PV = 13mnu2. We know that pressure depends on volume, number of molecules and root mean square velocity. Also, root mean square velocity depends on the temperature of the gas. Since the number of molecules, volume and temperature are constant, pressure of the gas will not change. 

Thus, (c) is the correct answer.

Answer:


Average momentum of a gas sample is zero, so it does not depend upon any of these parameters.

Thus, (d) is the correct answer.

Page No 34:

Question 6:


Average momentum of a gas sample is zero, so it does not depend upon any of these parameters.

Thus, (d) is the correct answer.

Answer:

(a) The kinetic energy of 1 mole
(c) The number of molecules in 1 mole

Kinetic energy per mole of an ideal gas is directly proportional to T. So, it will be the same for all ideal gases. 
Number of molecules in 1 mole of an ideal is the same for all ideal gases because ideal gases obey Avogadro's law.
Thus, (a) and (c) are correct answers.

Page No 34:

Question 7:

(a) The kinetic energy of 1 mole
(c) The number of molecules in 1 mole

Kinetic energy per mole of an ideal gas is directly proportional to T. So, it will be the same for all ideal gases. 
Number of molecules in 1 mole of an ideal is the same for all ideal gases because ideal gases obey Avogadro's law.
Thus, (a) and (c) are correct answers.

Answer:


In an ideal gas, the equation of state is given byPV=nRTPV=nNARNATPV=nNAkT1nNA=kTPVMultiplying both sides by mass of the gas M, we getMnNA=MkTPVNow, nNA gives the total number of molecules of the gas.Also, MnNA gives the mass of a single molecule.Hence,MkTPV is the mass of a single molecule of the gas, Molecular mass is a property of the gas. 

Thus, (d) is the correct answer.

Page No 34:

Question 1:


In an ideal gas, the equation of state is given byPV=nRTPV=nNARNATPV=nNAkT1nNA=kTPVMultiplying both sides by mass of the gas M, we getMnNA=MkTPVNow, nNA gives the total number of molecules of the gas.Also, MnNA gives the mass of a single molecule.Hence,MkTPV is the mass of a single molecule of the gas, Molecular mass is a property of the gas. 

Thus, (d) is the correct answer.

Answer:

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.  
Pressure, P = 1.01325×105 Pa    
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we  get

PV = nRT

V = RTP 

V=8.314×2731.01325×105=0.0224 m3

   

Page No 34:

Question 2:

Here,
STP means a system having a temperature of 273 K and 1 atm pressure.  
Pressure, P = 1.01325×105 Pa    
No of moles, n = 1 mol
Temperature, T = 273 K

Applying the equation of an ideal gas, we  get

PV = nRT

V = RTP 

V=8.314×2731.01325×105=0.0224 m3

   

Answer:

Here,
Volume of ideal gas at STP = 22.4 L
Number of molecules in 22.4 L of ideal gas at STP = 6.022×1023
 Number of molecules in 22.4​×103 cmof ideal gas at STP = ​6.022×1023

Now,
Number of molecules in 1 cm3 of ideal gas at STP = 6.022×102322.4×103=2.688×1019 

Page No 34:

Question 3:

Here,
Volume of ideal gas at STP = 22.4 L
Number of molecules in 22.4 L of ideal gas at STP = 6.022×1023
 Number of molecules in 22.4​×103 cmof ideal gas at STP = ​6.022×1023

Now,
Number of molecules in 1 cm3 of ideal gas at STP = 6.022×102322.4×103=2.688×1019 

Answer:

Given:
Volume of ideal gas, V = 1 cm3= 10-6 m​3
Temperature of ideal gas, T = 0 °C = 273 K
Pressure of mercury, P = 10−8 m of Hg
Density of ideal gas, ρ = 13600 kgm-3
Pressure P is given by
P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,

Using the ideal gas equation, we get
          
n=PVRTn=ρgh×VRTn=13600×9.8×10-8×10-68.31×273n=5.87×10-13

Number of molecules = N × n
                                    = 6.023 × 1023 × 5.874 × 10−13
                                    = 35.384 × 1010
                                    = 3.538 × 1011

Page No 34:

Question 4:

Given:
Volume of ideal gas, V = 1 cm3= 10-6 m​3
Temperature of ideal gas, T = 0 °C = 273 K
Pressure of mercury, P = 10−8 m of Hg
Density of ideal gas, ρ = 13600 kgm-3
Pressure P is given by
P = ρgh
Here,
ρ = density of ideal gas
g = acceleration due to gravity,

Using the ideal gas equation, we get
          
n=PVRTn=ρgh×VRTn=13600×9.8×10-8×10-68.31×273n=5.87×10-13

Number of molecules = N × n
                                    = 6.023 × 1023 × 5.874 × 10−13
                                    = 35.384 × 1010
                                    = 3.538 × 1011

Answer:

We know that 22.4 L of O2 contains 1 mol O2 at STP. Thus,

22.4×103 cm3 of O2=1 mol O21 cm3 of O2 = 122.4×103  mol O21 mol of O2 = 32 g122.4×103 mol of O2 =3222.4×103 =1.43×10-3 g                                                                 = 1.43 mg      

Page No 34:

Question 5:

We know that 22.4 L of O2 contains 1 mol O2 at STP. Thus,

22.4×103 cm3 of O2=1 mol O21 cm3 of O2 = 122.4×103  mol O21 mol of O2 = 32 g122.4×103 mol of O2 =3222.4×103 =1.43×10-3 g                                                                 = 1.43 mg      

Answer:

Let the pressure and temperature for the vessels of volume V0 and 2V0 be P​1, T1 and P2 , T2, respectively.
Since the two vessels have the same mass of gas, n1 = n2 = n.
T1=300 KT2 = 600 KUsing the equation of state for perfect gas, we getPV=nRTFor the vessel of volume Vo:P1Vo=nRT1                                   ...1For the vessel of volume 2Vo:P22Vo=nRT2                               ...2Dividing eq. 2 by eq. 1, we get2P2P1=T2T1=600300=2P2P1=1P2:P1=1:1

Page No 34:

Question 6:

Let the pressure and temperature for the vessels of volume V0 and 2V0 be P​1, T1 and P2 , T2, respectively.
Since the two vessels have the same mass of gas, n1 = n2 = n.
T1=300 KT2 = 600 KUsing the equation of state for perfect gas, we getPV=nRTFor the vessel of volume Vo:P1Vo=nRT1                                   ...1For the vessel of volume 2Vo:P22Vo=nRT2                               ...2Dividing eq. 2 by eq. 1, we get2P2P1=T2T1=600300=2P2P1=1P2:P1=1:1

Answer:

Given:
Volume of electric bulb, V = 250 cc
Temperature at which manufacturing takes place, T = 27  + 273  = 300 K
Height of mercury, h = 10−3 mm
Density of mercury, ρ =  13600 kgm−3
Avogadro constant, N = 6 × 1023 mol−1
Pressure P is given by
P = ρgh

Using the ideal gas equation, we get
PV=nRT

PV = nRTn =PVRTn=ρghVRTn =10-6×13600×10×250×10-6 8.314×300Now, number of molecules =nN=10-6×13600×10×250×10-6 8.314×300×6×1023=8×1015  

Page No 34:

Question 7:

Given:
Volume of electric bulb, V = 250 cc
Temperature at which manufacturing takes place, T = 27  + 273  = 300 K
Height of mercury, h = 10−3 mm
Density of mercury, ρ =  13600 kgm−3
Avogadro constant, N = 6 × 1023 mol−1
Pressure P is given by
P = ρgh

Using the ideal gas equation, we get
PV=nRT

PV = nRTn =PVRTn=ρghVRTn =10-6×13600×10×250×10-6 8.314×300Now, number of molecules =nN=10-6×13600×10×250×10-6 8.314×300×6×1023=8×1015  

Answer:

Given:
Maximum pressure that the cylinder can bear, Pmax = 1.0 × 106 Pa
Pressure in the gas cylinder, P1 = 8.0 × 105 Pa
Temperature in the cylinder, T1 = 300 K

Let T2 be the temperature at which the cylinder will break.
Volume is constant. Thus,                  (Given)
V1= V2 = V

Applying the five variable gas equation, we get

P1VT1=P2VT2     (V1=V2)P1T1=P2T2 T2=P2×T1P1T2=1.0×106×3008.0×105=375 K

Page No 34:

Question 8:

Given:
Maximum pressure that the cylinder can bear, Pmax = 1.0 × 106 Pa
Pressure in the gas cylinder, P1 = 8.0 × 105 Pa
Temperature in the cylinder, T1 = 300 K

Let T2 be the temperature at which the cylinder will break.
Volume is constant. Thus,                  (Given)
V1= V2 = V

Applying the five variable gas equation, we get

P1VT1=P2VT2     (V1=V2)P1T1=P2T2 T2=P2×T1P1T2=1.0×106×3008.0×105=375 K

Answer:

Given:
Mass of hydrogen, m = 2 g
Volume of the vessel, V = 0.02 m3
Temperature in the vessel, T = 300 K
Molecular mass of the hydrogen, M = 2 u

No of moles, n = mM=22= 1 mole

Rydberg's constant, R = 8.3 J/Kmol

From the ideal gas equation, we get

PV = nRT
P=nRTVP=1×8.3×3000.02
P = 1.24 × 105 Pa

Page No 34:

Question 9:

Given:
Mass of hydrogen, m = 2 g
Volume of the vessel, V = 0.02 m3
Temperature in the vessel, T = 300 K
Molecular mass of the hydrogen, M = 2 u

No of moles, n = mM=22= 1 mole

Rydberg's constant, R = 8.3 J/Kmol

From the ideal gas equation, we get

PV = nRT
P=nRTVP=1×8.3×3000.02
P = 1.24 × 105 Pa

Answer:

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas, ρ= 1.25 × 10−3 gcm−3 =1.25 kgm−3
Pressure, P = 1.01325 × 105 Pa   (At STP)
Temperature, T = 273 K    (At STP)

Using the ideal gas equation, we get

PV = nRT                    ...(1)n = mM                        ...(2) PV = mMRTM=mVRTPM=ρRTPM=1.25×8.31×273105M=2.83×10-2          =28.3 g-mol-1 

Page No 34:

Question 10:

Let:
m = Mass of the gas
M = Molecular mass of the gas
Now,
Density of ideal gas, ρ= 1.25 × 10−3 gcm−3 =1.25 kgm−3
Pressure, P = 1.01325 × 105 Pa   (At STP)
Temperature, T = 273 K    (At STP)

Using the ideal gas equation, we get

PV = nRT                    ...(1)n = mM                        ...(2) PV = mMRTM=mVRTPM=ρRTPM=1.25×8.31×273105M=2.83×10-2          =28.3 g-mol-1 

Answer:

Here,
Temperature in Simla, T1= 15 + 273 = 288 K       
Pressure in Simla, P​1 = 0.72  m of Hg 
Temperature in Kalka, T2= 35+273 = 308 K
Pressure in Kalka, P​2 = 0.76  m of Hg 
Let density of air
at Simla and Kalka be ρ1 and ρ2, respectively. Then,
PV=mMRTmV=PMRTρ=PMRT

Thus,

ρ1=P1MRT1
ρ2=P2MRT2

Taking ratios, we get

ρ1ρ2=P1T1×T2P2ρ1ρ2=0.72288×3080.76ρ2ρ1=0.987

Page No 34:

Question 11:

Here,
Temperature in Simla, T1= 15 + 273 = 288 K       
Pressure in Simla, P​1 = 0.72  m of Hg 
Temperature in Kalka, T2= 35+273 = 308 K
Pressure in Kalka, P​2 = 0.76  m of Hg 
Let density of air
at Simla and Kalka be ρ1 and ρ2, respectively. Then,
PV=mMRTmV=PMRTρ=PMRT

Thus,

ρ1=P1MRT1
ρ2=P2MRT2

Taking ratios, we get

ρ1ρ2=P1T1×T2P2ρ1ρ2=0.72288×3080.76ρ2ρ1=0.987

Answer:

Since the separator initially divides the cylinder equally, the number of moles of gas are equal in the two parts. Thus,
n= n2= n

​Volume of the first part = V
Volume of the second part =​3V

It is given that the walls are diathermic. So, temperature of the two parts is equal. Thus,
T1 = T2 = T
Let pressure of first and second parts be P1 and P2, respectively.
For first part:Applying equation of state, we getP1V=nRT                                        ...1For second part:Applying equation of state, we getP23V=nRT                                   ...2    Dividing eq. 1 by eq. 2, we getP1VP23V=1P1P2=31P1:P2=3:1

Page No 34:

Question 12:

Since the separator initially divides the cylinder equally, the number of moles of gas are equal in the two parts. Thus,
n= n2= n

​Volume of the first part = V
Volume of the second part =​3V

It is given that the walls are diathermic. So, temperature of the two parts is equal. Thus,
T1 = T2 = T
Let pressure of first and second parts be P1 and P2, respectively.
For first part:Applying equation of state, we getP1V=nRT                                        ...1For second part:Applying equation of state, we getP23V=nRT                                   ...2    Dividing eq. 1 by eq. 2, we getP1VP23V=1P1P2=31P1:P2=3:1

Answer:

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, M​0 = 2 g/mol=0.002 kg /mol
We know,
C=3RTM0C=3×8.3×3000.002C=1932.6 ms-1

In the second case, let the required temperature be T.
​
Applying the same formula, we get

3×8.3T0.002=2×1932.6T=1200 K

Page No 34:

Question 13:

Here,
Temperature of hydrogen gas, T = 300 K
Molar mass of hydrogen, M​0 = 2 g/mol=0.002 kg /mol
We know,
C=3RTM0C=3×8.3×3000.002C=1932.6 ms-1

In the second case, let the required temperature be T.
​
Applying the same formula, we get

3×8.3T0.002=2×1932.6T=1200 K

Answer:

Here,
V = 10-3 m3
Density = 0.177 kgm-3
P = 105pa   
C=3Pρ=3×1050.177                    =1301.9 ms-1

Page No 34:

Question 14:

Here,
V = 10-3 m3
Density = 0.177 kgm-3
P = 105pa   
C=3Pρ=3×1050.177                    =1301.9 ms-1

Answer:

We know from kinetic theory of gases that the average translational energy per molecule is 32kT.
Now,
Eavg= 0.040 eV = 0.040×1.6×10-19=6.4×10-21J 
6.40×10-21=32×1.38×10-23×TT=23×6.40×10-211.38×10-23=309.2 K

Page No 34:

Question 15:

We know from kinetic theory of gases that the average translational energy per molecule is 32kT.
Now,
Eavg= 0.040 eV = 0.040×1.6×10-19=6.4×10-21J 
6.40×10-21=32×1.38×10-23×TT=23×6.40×10-211.38×10-23=309.2 K

Answer:

Here,Vavg=8RTπM=8×8.83×3003.14×0.032=445.25 m/sWe know,T=DistanceSpeed=6400000×2445.25=28747.83 h3600=7.985 h=8 h

Page No 34:

Question 16:

Here,Vavg=8RTπM=8×8.83×3003.14×0.032=445.25 m/sWe know,T=DistanceSpeed=6400000×2445.25=28747.83 h3600=7.985 h=8 h

Answer:

Here,
​m = 6.64 × 10−27 kg
T = 273 K

Average speed of the He atom is given byVavg=8kTπm=8×1.38×10-23×2733.14×6.64×10-27=1202.31

We know,
Momentum = m × Vavg
= 6.64 × 10−27 × 1201.35
= 7.97 × 10−24
= 8 × 10−24 kg-m/s

Page No 34:

Question 17:

Here,
​m = 6.64 × 10−27 kg
T = 273 K

Average speed of the He atom is given byVavg=8kTπm=8×1.38×10-23×2733.14×6.64×10-27=1202.31

We know,
Momentum = m × Vavg
= 6.64 × 10−27 × 1201.35
= 7.97 × 10−24
= 8 × 10−24 kg-m/s

Answer:

Mean velocity is given by

Vavg=8RTπM

Let temperature for H and He respectively be  T​1 and T2, respectively.
For hydrogen:
MH = 2g = 2×10-3 kg
For helium:
MHe= 4 g = 4×10-3 kg

Now,
A/q8RT1πMH=8RT2πMHe8RT12×10-3π=8RT2π×4×10-3T12=T24T1T2=12T1:T2=1:2

Page No 34:

Question 18:

Mean velocity is given by

Vavg=8RTπM

Let temperature for H and He respectively be  T​1 and T2, respectively.
For hydrogen:
MH = 2g = 2×10-3 kg
For helium:
MHe= 4 g = 4×10-3 kg

Now,
A/q8RT1πMH=8RT2πMHe8RT12×10-3π=8RT2π×4×10-3T12=T24T1T2=12T1:T2=1:2

Answer:

Mean speed of the molecule is given by

8RTπMFor H molecule, M =2×10-3kg =4RT×103π

For escape velocity of Earth:

 Let r be the radius of Earth.v=2GMrMultiplying neumerator and denominator by R, we getvc=GMr22rg=GMr2vc=2gr

4RT×103π=2gr2×8.314×T×1033.142=9.8×6.37×106T11800 K



Page No 35:

Question 19:

Mean speed of the molecule is given by

8RTπMFor H molecule, M =2×10-3kg =4RT×103π

For escape velocity of Earth:

 Let r be the radius of Earth.v=2GMrMultiplying neumerator and denominator by R, we getvc=GMr22rg=GMr2vc=2gr

4RT×103π=2gr2×8.314×T×1033.142=9.8×6.37×106T11800 K

Answer:

We know,

Vavg=8RTπM

Molar mass of H2 = MH = 2×10-3 kg

Molar mass of N2 = MN = 28×10-3 kg
Now,
<V>H=8RTπMH<V>N=8RTπMN<V>H<V>N=MNMH=282=14=3.74

Page No 35:

Question 20:

We know,

Vavg=8RTπM

Molar mass of H2 = MH = 2×10-3 kg

Molar mass of N2 = MN = 28×10-3 kg
Now,
<V>H=8RTπMH<V>N=8RTπMN<V>H<V>N=MNMH=282=14=3.74

Answer:

Let the temperature of gas in both the chambers be T.
Let the molar mass of gas in the left chamber and right chamber be M​1 and M2, respectively.
Let mass of gas in the left and right chamber be m1 and m2, respectively. Then,

A/Qvrms=3kTm1=8kTπm2        3m1 = 8m23.14        m2m1=1.18

Page No 35:

Question 21:

Let the temperature of gas in both the chambers be T.
Let the molar mass of gas in the left chamber and right chamber be M​1 and M2, respectively.
Let mass of gas in the left and right chamber be m1 and m2, respectively. Then,

A/Qvrms=3kTm1=8kTπm2        3m1 = 8m23.14        m2m1=1.18

Answer:

Here,
λ=1.38×10-8 m
T = 273 K
M = 2×10-3 kg

Average speed of the H molecules is given  by

vavg=8RTπM=8×8.31×2733.14×2×10-3=1700 ms-1

The time between two collisions is given by

t=λvavgt=1.38×10-81700t=8×10-12 sNumber of collisions in 1 s = 18.11×10-12=1.23×1011 

Page No 35:

Question 22:

Here,
λ=1.38×10-8 m
T = 273 K
M = 2×10-3 kg

Average speed of the H molecules is given  by

vavg=8RTπM=8×8.31×2733.14×2×10-3=1700 ms-1

The time between two collisions is given by

t=λvavgt=1.38×10-81700t=8×10-12 sNumber of collisions in 1 s = 18.11×10-12=1.23×1011 

Answer:


Here,
P = 105 Pa
T = 300 K
For H2
M = 2×10-3 kg

a Mean speed is given by<v>=8RTπM=8×8.3×300×72×103×22=1780 ms1Let us consider a cubic volume of 1 m3.V=1 m3Momentum of 1 molecule normal to the striking surface before collision = musin450Momentum of 1 molecule normal to the striking surface after collision = musin450Change in momentum of the molecule = 2musin450=2muChange in momentum of n molecules = 2mnusin450=2mnuLet Δt be the time taken in changing the momentum.Force per unit area due to one molecule = 2muΔt=2muΔtObserved pressure due to collision by n molecules =2mnuΔt=105n=2mnuΔt2muΔt=1052mu6.0×1023 molecules = 2×103kg1 molecule = 2×1036×1023=3.3×1027kgn=1052×3.3×1027×1780=1.2×1028

Page No 35:

Question 23:


Here,
P = 105 Pa
T = 300 K
For H2
M = 2×10-3 kg

a Mean speed is given by<v>=8RTπM=8×8.3×300×72×103×22=1780 ms1Let us consider a cubic volume of 1 m3.V=1 m3Momentum of 1 molecule normal to the striking surface before collision = musin450Momentum of 1 molecule normal to the striking surface after collision = musin450Change in momentum of the molecule = 2musin450=2muChange in momentum of n molecules = 2mnusin450=2mnuLet Δt be the time taken in changing the momentum.Force per unit area due to one molecule = 2muΔt=2muΔtObserved pressure due to collision by n molecules =2mnuΔt=105n=2mnuΔt2muΔt=1052mu6.0×1023 molecules = 2×103kg1 molecule = 2×1036×1023=3.3×1027kgn=1052×3.3×1027×1780=1.2×1028

Answer:

Here,P1=2×105 PaP2=?T1=293 KT2=313 KV2=V1+0.02V1=V1(1.02)Now,P1V1T1=P2V2T22×105V1293=P2V1(1.02)313P2=2×105×313293×1.02=209 kPa

Page No 35:

Question 24:

Here,P1=2×105 PaP2=?T1=293 KT2=313 KV2=V1+0.02V1=V1(1.02)Now,P1V1T1=P2V2T22×105V1293=P2V1(1.02)313P2=2×105×313293×1.02=209 kPa

Answer:

Here,V1=1.0×103m3T1=400KP1=1.5×105 PaP2=1.0×105 PaT2=300M=32 gNumber of moles in the jar before n1=P1V1RT1Volume of the gas when pressure becomes equal to external pressure is given byP1V1T1=P2V2T2V2=P1V1T2P2T1V2=1.5×105×1.0×103×3001.0×105×400=1.125×103Net volume of leaked gas = V2V1=1.125×1031.0×103=1.25×104m3Let n2 be the number of moles of leaked gas. Applying equation of state on this amount of gas, we get n2=P2V2RT2=1.0×105×1.25×1048.3×300=0.005Mass of leaked gas= 32×0.005=0.16 g

Page No 35:

Question 25:

Here,V1=1.0×103m3T1=400KP1=1.5×105 PaP2=1.0×105 PaT2=300M=32 gNumber of moles in the jar before n1=P1V1RT1Volume of the gas when pressure becomes equal to external pressure is given byP1V1T1=P2V2T2V2=P1V1T2P2T1V2=1.5×105×1.0×103×3001.0×105×400=1.125×103Net volume of leaked gas = V2V1=1.125×1031.0×103=1.25×104m3Let n2 be the number of moles of leaked gas. Applying equation of state on this amount of gas, we get n2=P2V2RT2=1.0×105×1.25×1048.3×300=0.005Mass of leaked gas= 32×0.005=0.16 g

Answer:

Here,V1=43π(2.0×103)3h=3.3 mP1=Po+ρghP1=1.0×105+1000×9.8×3.3P1=1.32×105 PaP2=1.0×105 PaSince temperature remains the same, applying Boyle's law we getP1V1=P2V2V2=P1V1P2V2=1.32×105×43π(2.0×103)31.0×105Let R2 be the new radius. Then,43πR23=1.32×105×43π(2.0×103)31.0×105R23=1.32×105×(2.0×103)31.0×105R3=1.32×105×(2.0×103)31.0×1053R3=2.2×103m

Page No 35:

Question 26:

Here,V1=43π(2.0×103)3h=3.3 mP1=Po+ρghP1=1.0×105+1000×9.8×3.3P1=1.32×105 PaP2=1.0×105 PaSince temperature remains the same, applying Boyle's law we getP1V1=P2V2V2=P1V1P2V2=1.32×105×43π(2.0×103)31.0×105Let R2 be the new radius. Then,43πR23=1.32×105×43π(2.0×103)31.0×105R23=1.32×105×(2.0×103)31.0×105R3=1.32×105×(2.0×103)31.0×1053R3=2.2×103m

Answer:

Here,P1=2×105paV1=0.002m3V2=0.0005m3T1=T2=300KNumber of moles initially, n1=P1V1RT1n1=2×105×0.0028.3×300n1=0.16Applying equation of state, we get P2V2=n2RTAssuming the final pressure becomes equal to the atmospheric pressure, we getP2=1.0×105pan2=P2V2RTn2=1.0×105×0.00058.3×300n2=0.02Number of leaked moles = n2n1                                      =0.160.02                                      =0.14

Page No 35:

Question 27:

Here,P1=2×105paV1=0.002m3V2=0.0005m3T1=T2=300KNumber of moles initially, n1=P1V1RT1n1=2×105×0.0028.3×300n1=0.16Applying equation of state, we get P2V2=n2RTAssuming the final pressure becomes equal to the atmospheric pressure, we getP2=1.0×105pan2=P2V2RTn2=1.0×105×0.00058.3×300n2=0.02Number of leaked moles = n2n1                                      =0.160.02                                      =0.14

Answer:

Here,m=0.040gM=4gn=0.0404=0.01T1=100+273 K=373 KHe is a monoatomic gas. Thus,Cv=3×(12R)Cv=1.5×8.3=12.45Let the initial internal energy be U1.Let the final internal energy be U2.U2U1=nCv(T2T1)0.01×12.45(T2373)=12T2=469 KThe temperature in 0C  can be obtained as follows: 469 - 273 = 1960 C

Page No 35:

Question 28:

Here,m=0.040gM=4gn=0.0404=0.01T1=100+273 K=373 KHe is a monoatomic gas. Thus,Cv=3×(12R)Cv=1.5×8.3=12.45Let the initial internal energy be U1.Let the final internal energy be U2.U2U1=nCv(T2T1)0.01×12.45(T2373)=12T2=469 KThe temperature in 0C  can be obtained as follows: 469 - 273 = 1960 C

Answer:


Applying equation of state of an ideal gas, we getPV=nRTP=nRTV                               ...1Taking differentials, we getPdV+VdP=nRdT               ...2Applying the additional law, we getPV2=cV2dP+2VPdV=0VdP+2PdV=0                    ...3Subtracting eq. 3 from eq. 2, we getPdV=nRdTdV=nRPdTNow,dV=VTdT                From eq. 1dVV=dTTIntegrating between T2 and T1, we getV12V=T1T2ln(2V)ln(V)=ln(T1)ln(T2)ln(2VV)=ln(T1T2)T2=T12

Page No 35:

Question 29:


Applying equation of state of an ideal gas, we getPV=nRTP=nRTV                               ...1Taking differentials, we getPdV+VdP=nRdT               ...2Applying the additional law, we getPV2=cV2dP+2VPdV=0VdP+2PdV=0                    ...3Subtracting eq. 3 from eq. 2, we getPdV=nRdTdV=nRPdTNow,dV=VTdT                From eq. 1dVV=dTTIntegrating between T2 and T1, we getV12V=T1T2ln(2V)ln(V)=ln(T1)ln(T2)ln(2VV)=ln(T1T2)T2=T12

Answer:

Here, V = 0.166 m3T=300 KMass of O2=1.60 gMO=32 gnO=1.6032=0.05Mass of N2=2.80 gMN=28 gnN=2.8028=0.1Partial pressure of O2 is given byPO=nORTV=0.05×8.3×3000.166=750Partial pressure of N2 is given byPN=nNRTV=0.1×8.3×3000.166=1500Total pressure is sum of the partial pressures.P=PN+PO=750+1500=2250 Pa

Page No 35:

Question 30:

Here, V = 0.166 m3T=300 KMass of O2=1.60 gMO=32 gnO=1.6032=0.05Mass of N2=2.80 gMN=28 gnN=2.8028=0.1Partial pressure of O2 is given byPO=nORTV=0.05×8.3×3000.166=750Partial pressure of N2 is given byPN=nNRTV=0.1×8.3×3000.166=1500Total pressure is sum of the partial pressures.P=PN+PO=750+1500=2250 Pa

Answer:

Here,h=1 mP1=0.75mHg = 0.75ρg Paρ=13500 kg/m3Let h be the height of the mercury above the piston.P2=P1+hρgLet the CSA be A.V1=Ah=AV2=(1h)AApplying Boyle's law, we getP1V1=P2V20.75ρgA=P2(1h)A0.75ρg=(0.75ρg+hρg)(1h)0.75=(0.75+h)(1h)h=0.25 mh = 25 cm

Page No 35:

Question 31:

Here,h=1 mP1=0.75mHg = 0.75ρg Paρ=13500 kg/m3Let h be the height of the mercury above the piston.P2=P1+hρgLet the CSA be A.V1=Ah=AV2=(1h)AApplying Boyle's law, we getP1V1=P2V20.75ρgA=P2(1h)A0.75ρg=(0.75ρg+hρg)(1h)0.75=(0.75+h)(1h)h=0.25 mh = 25 cm

Answer:

Let the partial pressure of the gas in chamber A and B be PA' and PB', respectively.Applying equation of state for gas A, we getPAVTA=PA'2VTPA'=PAT2TASimilarly, for gas B:PB'=PBT2TBTotal pressure is the sum of the partial pressures. It is given byP=PA'+PB'=PAT2TA+PBT2TBP=T2(PATA+PBTB)PT=12(PATA+PBTB)

Page No 35:

Question 32:

Let the partial pressure of the gas in chamber A and B be PA' and PB', respectively.Applying equation of state for gas A, we getPAVTA=PA'2VTPA'=PAT2TASimilarly, for gas B:PB'=PBT2TBTotal pressure is the sum of the partial pressures. It is given byP=PA'+PB'=PAT2TA+PBT2TBP=T2(PATA+PBTB)PT=12(PATA+PBTB)

Answer:

a Here,V1=5×105m3P1=105 PaT1=273KM = 28.8 gP1V1=nRT1n=P1V1RT1mM=105×5×1058.3×273m=105×5×105×28.88.3×273m=0.0635 g

b Here,V1=5×105m3P1=105 PaP2=105 PaT1=273KT2=373KM = 28.8 gP1V1T1=P2V2T2 5×105273=V2373V2= 5×105×373273V2=6.831×105Volume of expelled air = 6.831×1055×105=1.831×105Applying equation of state, we getPV=nRTmM=PVRT=105×1.831×1058.3×373m=28.8×105×1.831×1058.3×373=0.017Thus, mass of expelled air = 0.017 gAmount of air in the container = 0.06350.017                                                    =0.0465

c Here,T=273KP=105PaV=5×105 m3Applying equation of state, we getPV=nRTP=nRTV=0.0465×8.3×27328.8×5×105P=0.731×10573kPa

Page No 35:

Question 33:

a Here,V1=5×105m3P1=105 PaT1=273KM = 28.8 gP1V1=nRT1n=P1V1RT1mM=105×5×1058.3×273m=105×5×105×28.88.3×273m=0.0635 g

b Here,V1=5×105m3P1=105 PaP2=105 PaT1=273KT2=373KM = 28.8 gP1V1T1=P2V2T2 5×105273=V2373V2= 5×105×373273V2=6.831×105Volume of expelled air = 6.831×1055×105=1.831×105Applying equation of state, we getPV=nRTmM=PVRT=105×1.831×1058.3×373m=28.8×105×1.831×1058.3×373=0.017Thus, mass of expelled air = 0.017 gAmount of air in the container = 0.06350.017                                                    =0.0465

c Here,T=273KP=105PaV=5×105 m3Applying equation of state, we getPV=nRTP=nRTV=0.0465×8.3×27328.8×5×105P=0.731×10573kPa

Answer:

Let the CSA of the tube be A.Initial volume of air, V1 = 20A cm = 0.2A Length of mercury, h = 0.1 m Let the pressure of the trapped air when the tube is inverted and vertical be P1.Now, pressure of the mercury and trapped air balances the atmospheric pressure. Thus,  P1+0.1ρg=0.75ρgP1=0.65ρgWhen the tube is inverted with the closed end down, the pressure acting upon the trapped air is given byAtmospheric pressure + Mercury column pressureNow,Pressure of trapped air = Atmospheric pressure + Mercury column pressure             In equilibriumP2=0.75ρg+0.1ρg=0.85ρgApplying the Boyle's law when the temperature remains constant, we getP1V1=P2V2Let the new height of the trapped air be x.0.65ρg0.2A = 0.85ρgxAx=0.15 m = 15 cm

Page No 35:

Question 34:

Let the CSA of the tube be A.Initial volume of air, V1 = 20A cm = 0.2A Length of mercury, h = 0.1 m Let the pressure of the trapped air when the tube is inverted and vertical be P1.Now, pressure of the mercury and trapped air balances the atmospheric pressure. Thus,  P1+0.1ρg=0.75ρgP1=0.65ρgWhen the tube is inverted with the closed end down, the pressure acting upon the trapped air is given byAtmospheric pressure + Mercury column pressureNow,Pressure of trapped air = Atmospheric pressure + Mercury column pressure             In equilibriumP2=0.75ρg+0.1ρg=0.85ρgApplying the Boyle's law when the temperature remains constant, we getP1V1=P2V2Let the new height of the trapped air be x.0.65ρg0.2A = 0.85ρgxAx=0.15 m = 15 cm

Answer:

Let CSA of the tube be A.On the colder side: P1=0.76 m HgT1=300KV1=V T2=273KV2=AxP1VT1=P2AxT2P2=P1VT2T1AxOn the hotter side:P1=0.76 m HgT1=300KV1'=V T2'=400KV2'=AyP1'VT1=P2'AyT2'P2'=P1VT2'T1AyIn equilibrium, the pressures on both side will balance each other.P2'=P2P1VT2'T1Ay=P1VT2T1AxT2'y=T2xFrom the length of the tube, we getx+y+0.1=1y=0.9x400(0.9x)=273xx=0.365 m x=36.5 cm

Page No 35:

Question 35:

Let CSA of the tube be A.On the colder side: P1=0.76 m HgT1=300KV1=V T2=273KV2=AxP1VT1=P2AxT2P2=P1VT2T1AxOn the hotter side:P1=0.76 m HgT1=300KV1'=V T2'=400KV2'=AyP1'VT1=P2'AyT2'P2'=P1VT2'T1AyIn equilibrium, the pressures on both side will balance each other.P2'=P2P1VT2'T1Ay=P1VT2T1AxT2'y=T2xFrom the length of the tube, we getx+y+0.1=1y=0.9x400(0.9x)=273xx=0.365 m x=36.5 cm

Answer:


Here,Initial pressure = Atmospheric pressure + Pressure due to mercury P1=Po+PHgLet the CSA of the tube be A.P1=0.76+0.2 =0.96 m HgT1=T2=TV1=0.43 AIf the tube is slanted, then the atmospheric pressure Poremains the same. Only the PHg changesP2=Po+PHgcos600=0.76+0.2×0.5=0.86 P1V1=P2V2V2=P1V1P2=0.96×0.43A0.86 Let the length of the air column be l.Al=P1V1P2=0.96×0.43A0.86 l=0.48 ml=48 cm



Page No 36:

Question 36:


Here,Initial pressure = Atmospheric pressure + Pressure due to mercury P1=Po+PHgLet the CSA of the tube be A.P1=0.76+0.2 =0.96 m HgT1=T2=TV1=0.43 AIf the tube is slanted, then the atmospheric pressure Poremains the same. Only the PHg changesP2=Po+PHgcos600=0.76+0.2×0.5=0.86 P1V1=P2V2V2=P1V1P2=0.96×0.43A0.86 Let the length of the air column be l.Al=P1V1P2=0.96×0.43A0.86 l=0.48 ml=48 cm

Answer:

Let the initial pressure of the chambers A and B be PA1 and PB1, respectively.
Let the final pressure of chambers A and B be PA2 and PB2, respectively.

Let the CSA be A. VA1=0.20 ATA1=400 KVB1=0.1 ATB1=100 KAt first equilibrium, both side pressures will be the same. PA1=PB1Let the final temperature at equilibrium be T. Then,PA1VA1TA1=PA2VA2TPA10.2A400=PA2VA2TPA2=PA10.2AT400VA2                         ...1For second chamber:PB1VB1TB1=PB2VB2TPB10.1A100=PB2VB2TPB2=PB10.1AT100VB2                      ...2At second equilibrium, pressures on both sides will be the same.PA2=PB2PA10.2AT400VA2=PB10.1AT100VB2PA12VA2=PB1VB2VB2=2VA2                                  ...3Now,VB2+VA2=0.3A3VA2=0.3AVA2=0.1ALet VA2 be lA.l=0.1 m⇒ l=10 cm

Page No 36:

Question 37:

Let the initial pressure of the chambers A and B be PA1 and PB1, respectively.
Let the final pressure of chambers A and B be PA2 and PB2, respectively.

Let the CSA be A. VA1=0.20 ATA1=400 KVB1=0.1 ATB1=100 KAt first equilibrium, both side pressures will be the same. PA1=PB1Let the final temperature at equilibrium be T. Then,PA1VA1TA1=PA2VA2TPA10.2A400=PA2VA2TPA2=PA10.2AT400VA2                         ...1For second chamber:PB1VB1TB1=PB2VB2TPB10.1A100=PB2VB2TPB2=PB10.1AT100VB2                      ...2At second equilibrium, pressures on both sides will be the same.PA2=PB2PA10.2AT400VA2=PB10.1AT100VB2PA12VA2=PB1VB2VB2=2VA2                                  ...3Now,VB2+VA2=0.3A3VA2=0.3AVA2=0.1ALet VA2 be lA.l=0.1 m⇒ l=10 cm

Answer:

Let P be the pressure and n be the number of moles of gas inside the vessel at any given time t.  Suppose a small amount of gas of dn moles is pumped out and the decrease in pressure is dP.Applying equation of state to the gas inside the vessel, we get(PdP)Vo=(ndn)RTPVodPVo=nRTdnRTBut PVo=nRTVodP=dnRT                              ...1The pressure of the gas taken out is equal to the inner pressure.Applying equation of state, we get (PdP)dV =dnRTPdV=dnRT                                 ...2From eq. 1 and eq. 2, we getVodP=PdVdPP=dVVodVdt=rdV=rdtdV=rdt                                      ...3                 Since pressures decreases, rate is negativeNow,dPP=rdtVo                          From eq. 3   aIntegrating the equation P = P0 to P = P and time t = 0 to t = t, we getPoP=0tlnPlnPo=rtVoln(PPo)=rtVoP=PoertVo

bP = Po2Po2=PoertVoertVo=2rtVo=ln2t=Voln2r

Page No 36:

Question 38:

Let P be the pressure and n be the number of moles of gas inside the vessel at any given time t.  Suppose a small amount of gas of dn moles is pumped out and the decrease in pressure is dP.Applying equation of state to the gas inside the vessel, we get(PdP)Vo=(ndn)RTPVodPVo=nRTdnRTBut PVo=nRTVodP=dnRT                              ...1The pressure of the gas taken out is equal to the inner pressure.Applying equation of state, we get (PdP)dV =dnRTPdV=dnRT                                 ...2From eq. 1 and eq. 2, we getVodP=PdVdPP=dVVodVdt=rdV=rdtdV=rdt                                      ...3                 Since pressures decreases, rate is negativeNow,dPP=rdtVo                          From eq. 3   aIntegrating the equation P = P0 to P = P and time t = 0 to t = t, we getPoP=0tlnPlnPo=rtVoln(PPo)=rtVoP=PoertVo

bP = Po2Po2=PoertVoertVo=2rtVo=ln2t=Voln2r

Answer:

Given: p=po1+(VVo)2Multiplying both sides by V, we getpV=poV1+(VVo)2pV=RT                        From eq. 1Now,RT=poV1+(VVo)2T=1R(poVo1+(VoVo)2)           V=VoT=poVo2R

Page No 36:

Question 39:

Given: p=po1+(VVo)2Multiplying both sides by V, we getpV=poV1+(VVo)2pV=RT                        From eq. 1Now,RT=poV1+(VVo)2T=1R(poVo1+(VoVo)2)           V=VoT=poVo2R

Answer:

We know internal energy at a particular temperatureU=nCvTAir in home is are chiefly diatomic molecules, so Cv=52R U=5n2RTNow by eqn. of state nRT=PVU=52(nRT)=>U=52PVNow pressure P is constant also V of the room = constantThus,U=constant

Page No 36:

Question 40:

We know internal energy at a particular temperatureU=nCvTAir in home is are chiefly diatomic molecules, so Cv=52R U=5n2RTNow by eqn. of state nRT=PVU=52(nRT)=>U=52PVNow pressure P is constant also V of the room = constantThus,U=constant

Answer:

Here,P1=105 PaA=π(0.05)2L=0.2 mV=AL=0.0016 m3T1=300 KT2=600 Kμ=0.20Applying 5 variable equation of state, we getP1VT1=P2VT2P1T1=P2T2P2=T2T1×P1=600300×105P2=2×105Net pressure, P = P2P1 = 2×105105=105Total force acting on the stopper = PA =105×π×(0.05)2Applying law of friction, we getF=μN=0.2NN=Fμ=105×π×(0.05)20.2dNdl=N2πr=105×π×(0.05)20.2×2π×(0.05)=0.125×105dNdl=1.25×104​ N/m

Page No 36:

Question 41:

Here,P1=105 PaA=π(0.05)2L=0.2 mV=AL=0.0016 m3T1=300 KT2=600 Kμ=0.20Applying 5 variable equation of state, we getP1VT1=P2VT2P1T1=P2T2P2=T2T1×P1=600300×105P2=2×105Net pressure, P = P2P1 = 2×105105=105Total force acting on the stopper = PA =105×π×(0.05)2Applying law of friction, we getF=μN=0.2NN=Fμ=105×π×(0.05)20.2dNdl=N2πr=105×π×(0.05)20.2×2π×(0.05)=0.125×105dNdl=1.25×104​ N/m

Answer:


aSince pressure from outside and inside the cylinder is the same, there is no net pressureacting on the pistons. So, tension will be zerobT1=ToT2=2ToP1=po=105 PaCSA=ALet the pistons be L distance apart.V=ALApplying five variable gas equation, we getP1VT1=P2VT2105T0=P22ToP2=2×105=2PoNet force acting outside = 2P0P0=P0Force acting on a piston F= PoABy the free body diagram, we get F-T=0T = PoA

Page No 36:

Question 42:


aSince pressure from outside and inside the cylinder is the same, there is no net pressureacting on the pistons. So, tension will be zerobT1=ToT2=2ToP1=po=105 PaCSA=ALet the pistons be L distance apart.V=ALApplying five variable gas equation, we getP1VT1=P2VT2105T0=P22ToP2=2×105=2PoNet force acting outside = 2P0P0=P0Force acting on a piston F= PoABy the free body diagram, we get F-T=0T = PoA

Answer:

a Pressure of water above the water level of the bigger tank is given by P =(h2+ho)ρgLet the atmospheric pressure above the tube be Po.Total pressure above the tube = P0+P=(h2+ho)ρg+PoThis pressure initially is balanced by pressure above the tank 2Po.2Po=(h2+ho)ρg+Poh2=Poρghob Velocity of the efflux out of the outlet depends upon the total pressure above the outlet.Total pressure above the outlet = 2Po+(h1ho)ρgApplying Bernouli's law, we getLet the velocity of efflux be v1 and the velocity with which the level of the tank falls be v2.​ Pressure above the outlet is Po. Then, 2Po+(h1ho)ρgρ+gz+v222=Poρ+gz+v222Now, let the reference point of the liquid be the level of the outlet. Thus,z =0Po+(h1ho)ρgρ+v222=v122Again, the speed with which the water level of the tank goes down is very less compared to the velocity of the efflux. Thus,v2=0Po+(h1ho)ρgρ=v122v1=[2ρ(Po+(h1ho)ρg)]12
c)Water maintains its own level, so height of the water of the tankwill be h1when water will stop flowingThus height of water in the tube below the tank height will be = h1Hence height of the water above the tank height will be = h1

Page No 36:

Question 43:

a Pressure of water above the water level of the bigger tank is given by P =(h2+ho)ρgLet the atmospheric pressure above the tube be Po.Total pressure above the tube = P0+P=(h2+ho)ρg+PoThis pressure initially is balanced by pressure above the tank 2Po.2Po=(h2+ho)ρg+Poh2=Poρghob Velocity of the efflux out of the outlet depends upon the total pressure above the outlet.Total pressure above the outlet = 2Po+(h1ho)ρgApplying Bernouli's law, we getLet the velocity of efflux be v1 and the velocity with which the level of the tank falls be v2.​ Pressure above the outlet is Po. Then, 2Po+(h1ho)ρgρ+gz+v222=Poρ+gz+v222Now, let the reference point of the liquid be the level of the outlet. Thus,z =0Po+(h1ho)ρgρ+v222=v122Again, the speed with which the water level of the tank goes down is very less compared to the velocity of the efflux. Thus,v2=0Po+(h1ho)ρgρ=v122v1=[2ρ(Po+(h1ho)ρg)]12
c)Water maintains its own level, so height of the water of the tankwill be h1when water will stop flowingThus height of water in the tube below the tank height will be = h1Hence height of the water above the tank height will be = h1

Answer:

Atmospheric pressure inside the cylinderical vessel, P0=105 PaA=10cm2=10×104 m2Pressure due to the weight of the piston = mgA=1×9.810×104P1=105+9.8×103V1=0.2×10×104=2×104After evacution, external pressure above the piston =0P2=0+9.8×103Now,P1V1=P2V2Let L be the final length of the gas column. Then,V2=10×104(105+9.8×103)×0.2×10×104=9.8×103×10×104⇒L=2.2 m

Page No 36:

Question 44:

Atmospheric pressure inside the cylinderical vessel, P0=105 PaA=10cm2=10×104 m2Pressure due to the weight of the piston = mgA=1×9.810×104P1=105+9.8×103V1=0.2×10×104=2×104After evacution, external pressure above the piston =0P2=0+9.8×103Now,P1V1=P2V2Let L be the final length of the gas column. Then,V2=10×104(105+9.8×103)×0.2×10×104=9.8×103×10×104⇒L=2.2 m

Answer:

Here,V=50m3T=273+15=288KRH=40%a)Let x be the amount of water that evaporates.P=RH×SVP⇒P=0.4×1600=640PV=mMRTm=PVMRT=640×50×188.3×288=241gWater will evaporate until VP becomes equal to SVP.1600×50=m+xMRT1600×50=241+x18×8.3×288x361 gb)T=288+5=293KSVP=2400PaDifference in pressure = 24001600=800 PaLet x be the amount of water that evaporates.PV=xMRT800×50=x18×8.3×293x296 g

Page No 36:

Question 45:

Here,V=50m3T=273+15=288KRH=40%a)Let x be the amount of water that evaporates.P=RH×SVP⇒P=0.4×1600=640PV=mMRTm=PVMRT=640×50×188.3×288=241gWater will evaporate until VP becomes equal to SVP.1600×50=m+xMRT1600×50=241+x18×8.3×288x361 gb)T=288+5=293KSVP=2400PaDifference in pressure = 24001600=800 PaLet x be the amount of water that evaporates.PV=xMRT800×50=x18×8.3×293x296 g

Answer:


Here,P1=0.76 m HgP2=PT1=273 KT2=335KLet each of the bulbs have n1 moles initially.Let the number of moles left in second bulb after its pressure reached P be n2.Applying equation of state, we getP1Vn1T1=PVn2T20.76273n1=P335n2n2=273P335×0.76n1Number of moles left in the second bulb after the temperature rose = n1n2=n1273P335×0.76n1Let n3 moles be left when pressure reached P. Applying equation of state in the first bulb, we getP1Vn1T1=PVn3T10.76n1=Pn3n3=Pn10.76n3=its own n1 moles + the it recieved from the firstn3=n1+(n1n2)Pn10.76=n1+n1273P335×0.76n1P0.76=2273P335×0.76P=0.8375P=84 cm of Hg



Page No 37:

Question 46:


Here,P1=0.76 m HgP2=PT1=273 KT2=335KLet each of the bulbs have n1 moles initially.Let the number of moles left in second bulb after its pressure reached P be n2.Applying equation of state, we getP1Vn1T1=PVn2T20.76273n1=P335n2n2=273P335×0.76n1Number of moles left in the second bulb after the temperature rose = n1n2=n1273P335×0.76n1Let n3 moles be left when pressure reached P. Applying equation of state in the first bulb, we getP1Vn1T1=PVn3T10.76n1=Pn3n3=Pn10.76n3=its own n1 moles + the it recieved from the firstn3=n1+(n1n2)Pn10.76=n1+n1273P335×0.76n1P0.76=2273P335×0.76P=0.8375P=84 cm of Hg

Answer:

Here,
Relative humidity = 100% 
RH=Vapour pressure of airSVP at the same temperature=1Vapour pressure of air=SVP at the same temperatureSo, the air is saturated at 200C​. So, dew point is 200C. 

Page No 37:

Question 47:

Here,
Relative humidity = 100% 
RH=Vapour pressure of airSVP at the same temperature=1Vapour pressure of air=SVP at the same temperatureSo, the air is saturated at 200C​. So, dew point is 200C. 

Answer:

Here,T = 298KRH = 60%P = 1.04×105 paRH=Vapour pressure of water vapourSaturated vapour pressure     =0.6Saturated vapour pressure = 3.2×103 PaVapour pressure of water vapour (VP)= 0.6×3.2×103 =1.92×103 PaIf the water vapour is completely removed from the air, then net pressure =1.04×1051.92×103                                                                                                                              =1.02×105 Pa                                                                                                                                =102 kPa

Page No 37:

Question 48:

Here,T = 298KRH = 60%P = 1.04×105 paRH=Vapour pressure of water vapourSaturated vapour pressure     =0.6Saturated vapour pressure = 3.2×103 PaVapour pressure of water vapour (VP)= 0.6×3.2×103 =1.92×103 PaIf the water vapour is completely removed from the air, then net pressure =1.04×1051.92×103                                                                                                                              =1.02×105 Pa                                                                                                                                =102 kPa

Answer:

Here,Temperature = 20°CDew point = 10°CAir becomes saturated at 10°C. But if the room temperature is lowered to 15°C, the dew point will still be at 10°C.  

Page No 37:

Question 49:

Here,Temperature = 20°CDew point = 10°CAir becomes saturated at 10°C. But if the room temperature is lowered to 15°C, the dew point will still be at 10°C.  

Answer:

Here,RH=40%V1=10×106m3RH=VPSVP=0.4Let SVP = PoCondensation occurs when VP = Po.P1=0.4PoP2=PoSince the process is isothermal, applying Boyle's law we getP1V1=P2V2V2=P1V1P2V2=0.4Po×10×106PoV2=4.0×106V2=4.0cm3Thus water vapour condenses at volume 4.0 cm3.

Page No 37:

Question 50:

Here,RH=40%V1=10×106m3RH=VPSVP=0.4Let SVP = PoCondensation occurs when VP = Po.P1=0.4PoP2=PoSince the process is isothermal, applying Boyle's law we getP1V1=P2V2V2=P1V1P2V2=0.4Po×10×106PoV2=4.0×106V2=4.0cm3Thus water vapour condenses at volume 4.0 cm3.

Answer:

Here,Atmospheric pressure, P=0.76 m HgPressure due to water vapour inside, P' =0.754 mHgVapour pressure = PP'=0.760.754=0.006 mHgSVH = 0.01 mHgRH=Vapour pressureSVH×100%=0.0060.01×100%=60%

Page No 37:

Question 51:

Here,Atmospheric pressure, P=0.76 m HgPressure due to water vapour inside, P' =0.754 mHgVapour pressure = PP'=0.760.754=0.006 mHgSVH = 0.01 mHgRH=Vapour pressureSVH×100%=0.0060.01×100%=60%

Answer:


We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 760 mm. This gives the boiling point 650 of methyl alcohol.

For 0.5 atm pressure, corresponding pressure in mm Hg will be 375 mm. We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 375 mm. This gives the boiling point 480 of methyl alcohol.

Page No 37:

Question 52:


We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 760 mm. This gives the boiling point 650 of methyl alcohol.

For 0.5 atm pressure, corresponding pressure in mm Hg will be 375 mm. We drop a perpendicular on x-axis corresponding to the saturated vapour pressure 375 mm. This gives the boiling point 480 of methyl alcohol.

Answer:


Here,
T = 980F

When we convert the temperature to 0C, we get59(F32)=59(9832)=36.70C

We drop perpendicular corresponding to a temperature of 36.70C on Y-axis from the curve of pure water. This gives the boiling point of blood 50 mm of Hg. 

Page No 37:

Question 53:


Here,
T = 980F

When we convert the temperature to 0C, we get59(F32)=59(9832)=36.70C

We drop perpendicular corresponding to a temperature of 36.70C on Y-axis from the curve of pure water. This gives the boiling point of blood 50 mm of Hg. 

Answer:

Here,Dew point = 100C          ∵ Dew appears at 100CAt boiling point, SVP equals atmospheric pressure.At 200C, SVP = 17.5 mmHgAt dew point, SVP= 8.9 mmHgRH=SVP at dew pointSVP at air temperature×100%=8.917.5×100%=51%

Page No 37:

Question 54:

Here,Dew point = 100C          ∵ Dew appears at 100CAt boiling point, SVP equals atmospheric pressure.At 200C, SVP = 17.5 mmHgAt dew point, SVP= 8.9 mmHgRH=SVP at dew pointSVP at air temperature×100%=8.917.5×100%=51%

Answer:

We know that 1 m3 of air contains 30 g of water vapour at 300C. So, amount of water vapour in 50 m3 of air at 300C = 30×50 g=1500 gAlso, 1 m3 of air contains 16 g of water vapour at 200C. Amount of water vapour in 50 m3 of air at 200C = 16×50 g=800 gAmount of water vapour condensed= 1500  800 g=700 g

Page No 37:

Question 55:

We know that 1 m3 of air contains 30 g of water vapour at 300C. So, amount of water vapour in 50 m3 of air at 300C = 30×50 g=1500 gAlso, 1 m3 of air contains 16 g of water vapour at 200C. Amount of water vapour in 50 m3 of air at 200C = 16×50 g=800 gAmount of water vapour condensed= 1500  800 g=700 g

Answer:

Atmospheric pressure = 76 cm HgSVP = 0.80 cm HgWhen water is introduced into the barometer, water evaporates. Thus, it exerts its vapour pressure over the mercury meniscus.As more and more water evaporates, the vapour pressure increases thatforces down the mercury level further.Finally, when the volume is saturated with the vapour at the atmospheric temperature, the highest vapour pressure, i.e. SVP is observed and the fall of mercury levelreaches its minimum. Thus,Net pressure acting on the column = 76 0.80 cmHgNet length of Hg column at SVG = 75.2 cm  

Page No 37:

Question 56:

Atmospheric pressure = 76 cm HgSVP = 0.80 cm HgWhen water is introduced into the barometer, water evaporates. Thus, it exerts its vapour pressure over the mercury meniscus.As more and more water evaporates, the vapour pressure increases thatforces down the mercury level further.Finally, when the volume is saturated with the vapour at the atmospheric temperature, the highest vapour pressure, i.e. SVP is observed and the fall of mercury levelreaches its minimum. Thus,Net pressure acting on the column = 76 0.80 cmHgNet length of Hg column at SVG = 75.2 cm  

Answer:

Here,Atmospheric pressure, Po=99.4×103 PaSVP at 270C Pw = 3.4×103 PaT=300KV=50×106 m3Now,Pressure inside the jar = Pressure outside the jar           ∵ Level of water is same inside and outside of the jarPressure outside the jar = Atmospheric pressurePressure inside the jar = VP of oxygen + SVP of water at 270CP0=P+PwP=PoPw=99.4×1033.4×103=96×103Applying equation of state, we getPV=nRT96×103×50×106=n×8.3×300n=1.9277×103 mol 1.93×103 mol

Page No 37:

Question 57:

Here,Atmospheric pressure, Po=99.4×103 PaSVP at 270C Pw = 3.4×103 PaT=300KV=50×106 m3Now,Pressure inside the jar = Pressure outside the jar           ∵ Level of water is same inside and outside of the jarPressure outside the jar = Atmospheric pressurePressure inside the jar = VP of oxygen + SVP of water at 270CP0=P+PwP=PoPw=99.4×1033.4×103=96×103Applying equation of state, we getPV=nRT96×103×50×106=n×8.3×300n=1.9277×103 mol 1.93×103 mol

Answer:

Given:Let the CSA be A.Case 1:V1=(x74)ASVP = 1 cm HgAtmospheric pressure, Po=76 cm HgMercury column height = 74.0 cmLet P be the air pressure above the barometer. Then,Atmospheric pressure =SVP + Air pressure above the barometer mercury level + Mercury column height⇒1+P +74 =76 P = 1 cmCase 2:Atmospheric pressure, P0'=74.0 cm HgLet P' be the air pressure. Then,P'+72.10+1=76P'=0.9V2=(x72.1)AApplying Boyle's law, we get PV1=P'V2⇒1×(x74)A=0.9×(x72.1)Ax=91.1 cmLength of the tube = 91.1 cm 

Page No 37:

Question 58:

Given:Let the CSA be A.Case 1:V1=(x74)ASVP = 1 cm HgAtmospheric pressure, Po=76 cm HgMercury column height = 74.0 cmLet P be the air pressure above the barometer. Then,Atmospheric pressure =SVP + Air pressure above the barometer mercury level + Mercury column height⇒1+P +74 =76 P = 1 cmCase 2:Atmospheric pressure, P0'=74.0 cm HgLet P' be the air pressure. Then,P'+72.10+1=76P'=0.9V2=(x72.1)AApplying Boyle's law, we get PV1=P'V2⇒1×(x74)A=0.9×(x72.1)Ax=91.1 cmLength of the tube = 91.1 cm 

Answer:

Given: SVP at 0oC = 4.6 mm HgRH =40%VP of air (P1)SVP at the same temperature=0.4P14.6=0.4P1=1.84 mmHgHere, V is constant.T1=273KT2=293KApplying equationof state, we getP1VT1=P2VT2P2=P1T2T1P2=1.84×293273P2=1.97RH inside home=VP of air (P2)SVP at the same temperature×100%SVP at the same temperature=18mmHgRH=1.9718×100=10.9%

Page No 37:

Question 59:

Given: SVP at 0oC = 4.6 mm HgRH =40%VP of air (P1)SVP at the same temperature=0.4P14.6=0.4P1=1.84 mmHgHere, V is constant.T1=273KT2=293KApplying equationof state, we getP1VT1=P2VT2P2=P1T2T1P2=1.84×293273P2=1.97RH inside home=VP of air (P2)SVP at the same temperature×100%SVP at the same temperature=18mmHgRH=1.9718×100=10.9%

Answer:

Here,SVP=3600 PaT = 273 + 27 = 300KV = 1m3M = 18 g for waterRH=50%VPSVP=0.5VP=0.5×3600=1800Let m1 be the mass of water present in the 50% humid air.PV = nRTPV=m1MRT1800=m118×8.3×300m1=13gRequired pressure for saturation=3600 PaLet m2 be the amount of water required for saturation.3600=m2MRTm2=3600×188.3×300=26gTotal excess water vapour that has to be added = m2m1=3613=13g

Page No 37:

Question 60:

Here,SVP=3600 PaT = 273 + 27 = 300KV = 1m3M = 18 g for waterRH=50%VPSVP=0.5VP=0.5×3600=1800Let m1 be the mass of water present in the 50% humid air.PV = nRTPV=m1MRT1800=m118×8.3×300m1=13gRequired pressure for saturation=3600 PaLet m2 be the amount of water required for saturation.3600=m2MRTm2=3600×188.3×300=26gTotal excess water vapour that has to be added = m2m1=3613=13g

Answer:

Here,T=300KSVP=3300 Pa at 300 KRH=20%PSVP=0.2P=0.2×SVP=0.2×3300=660V=50 m3M=18 g Now,PV = nRT PV mMRT660×50=m18×8.3×300m=238.55 g≈238 g

Page No 37:

Question 61:

Here,T=300KSVP=3300 Pa at 300 KRH=20%PSVP=0.2P=0.2×SVP=0.2×3300=660V=50 m3M=18 g Now,PV = nRT PV mMRT660×50=m18×8.3×300m=238.55 g≈238 g

Answer:

Here,M=18g for waterm=500gV=50 m3T=300KSVP=3300PaRH=20%VPSVP=0.2VP=P1=0.2×3300=660PaPartial pressure P2 for evaporated water is given byP2V=mMRTP2=50018×50×8.31×300P2=1385PaTotal pressure, P=P1+P2=1385+660=2045PaRH=PSVP×100=20453300×100%=61.962%

Page No 37:

Question 62:

Here,M=18g for waterm=500gV=50 m3T=300KSVP=3300PaRH=20%VPSVP=0.2VP=P1=0.2×3300=660PaPartial pressure P2 for evaporated water is given byP2V=mMRTP2=50018×50×8.31×300P2=1385PaTotal pressure, P=P1+P2=1385+660=2045PaRH=PSVP×100=20453300×100%=61.962%

Answer:

(a) Relative humidity is given by

VPSVP at 15°C0.4=VP1.6×103
⇒ VP = 0.4 × 1.6 × 103

Evaporation occurs as long as the atmosphere is not saturated.

Net pressure change = 1.6 × 103 − 0.4 × 1.6 × 103
                                 = (1.6 − 0.4 × 1.6) 103
                                 = 0.96 × 103

Let the mass of water evaporated be m. Then,

0.96×103×50=m×8.3×28818m=0.96×50×18×1038.3×288=361.45361 g

(b) At 20°C, SVP = 2.4 KPa
At 15°C, SVP = 1.6 KPa

Net pressure change = (2.4 − 1.6) × 103 Pa
                                 = 0.8 × 103 Pa

Mass of water evaporated is given by

m=m'×8.3×29318m'=0.8×50×18×1038.3×293

= 296.06 ≈ 296 g



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