Kinetic Theory and Thermodynamics
LIST–I  LIST–II  
P.  In process I  1.  Work done by the gas is zero 
Q.  In process II  2.  Temperature of the gas remains unchanged 
R.  In process III  3.  No heat is exchanged between the gas and its surroundings 
S.  In process IV  4.  Work done by the gas is 6P_{0}V_{0} 
An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. 

Column1  Column2  Column3 
(I) ${\mathrm{W}}_{1\to 2}=\frac{1}{\mathrm{\gamma}1}\left({\mathrm{P}}_{2}{\mathrm{V}}_{2}{\mathrm{P}}_{1}{\mathrm{V}}_{1}\right)$  (i) Isothermal  
(II) ${\mathrm{W}}_{1\to 2}={\mathrm{PV}}_{2}+{\mathrm{PV}}_{1}$  (ii) Isochoric  
(III) ${\mathrm{W}}_{1\to 2}=0$  (iii) Isobaric  
(IV) ${\mathrm{W}}_{1\to 2}=\mathrm{nRT}\mathrm{In}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}$  (iv) Adiabatic 
Which of the following options is the only correct representation of a process in which ΔU = ΔQ – PΔV?
An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. 

Column1  Column2  Column3 
(I) ${\mathrm{W}}_{1\to 2}=\frac{1}{\mathrm{\gamma}1}\left({\mathrm{P}}_{2}{\mathrm{V}}_{2}{\mathrm{P}}_{1}{\mathrm{V}}_{1}\right)$  (i) Isothermal  
(II) ${\mathrm{W}}_{1\to 2}={\mathrm{PV}}_{2}+{\mathrm{PV}}_{1}$  (ii) Isochoric  
(III) ${\mathrm{W}}_{1\to 2}=0$  (iii) Isobaric  
(IV) ${\mathrm{W}}_{1\to 2}=\mathrm{nRT}\mathrm{In}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}$  (iv) Adiabatic 
Which one of the following options is the correct combination?
An ideal gas is undergoing a cyclic thermodynamics process in different ways as shown in the corresponding P–V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamics processes. Here γ is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas is n. 

Column1  Column2  Column3 
(I) ${\mathrm{W}}_{1\to 2}=\frac{1}{\mathrm{\gamma}1}\left({\mathrm{P}}_{2}{\mathrm{V}}_{2}{\mathrm{P}}_{1}{\mathrm{V}}_{1}\right)$  (i) Isothermal  
(II) ${\mathrm{W}}_{1\to 2}={\mathrm{PV}}_{2}+{\mathrm{PV}}_{1}$  (ii) Isochoric  
(III) ${\mathrm{W}}_{1\to 2}=0$  (iii) Isobaric  
(IV) ${\mathrm{W}}_{1\to 2}=\mathrm{nRT}\mathrm{In}\frac{{\mathrm{V}}_{2}}{{\mathrm{V}}_{1}}$  (iv) Adiabatic 
Which one of the following options correctly represents a thermodynamics process that is used as a correction in the determination of the speed of sound in an ideal gas?
One mole of a monatomic ideal gas is taken along two cyclic processes E→F→G→E and E→F→H→E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic.
Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

List I 

List II 
P. 
G→E 
1. 
160 P_{0}V_{0} ln2 
Q. 
G→H 
2. 
36 P_{0}V_{0} 
R. 
F→H 
3. 
24 P_{0}V_{0} 
S. 
F→G 
4. 
31 P_{0}V_{0} 
One mole of a monatomic ideal gas is taken through a cycle ABCDA as shown in the P −V diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.
Column I 
Column II 

(A) 
Process AB 
(P) 
Internal energy decreases 
(B) 
Process B C 
(Q) 
Internal energy increases 
(C) 
Process C D 
(R) 
Heat is lost. 
(D) 
Process D A 
(S) 
Heated is gained 
(T) 
Work is done on the gas 
(A)  (P)  ∆Q = –100 J  
(B)  (Q)  ∆Q = –50 J  
(C)  (R)  ∆Q = 100 J  
(D)  (S)  ∆Q = 30 J 
In which of the following options temperature increases?
(A)  (P)  ∆Q = –100 J  
(B)  (Q)  ∆Q = –50 J  
(C)  (R)  ∆Q = 100 J  
(D)  (S)  ∆Q = 30 J 
In which of the following options temperature decreases?
(A)  (P)  ∆Q = –100 J  
(B)  (Q)  ∆Q = –50 J  
(C)  (R)  ∆Q = 100 J  
(D)  (S)  ∆Q = 30 J 
In which case the final temperature T is maximum?
Use the following information to answer the next question.
A heat engine operating between 508 K and 400 K temperatures requires an electrical input of 2200 J for each cycle. 
The heat deposited in the cold reservoir in one cycle is_____________J.
Column I  Column II  Column III  
A.  n = 0  I.  Slope of PV curve is $\frac{P}{V}$  P.  Work done by gas this process is $\frac{5}{2}\left({P}_{1}{V}_{1}{P}_{2}{V}_{2}\right)$ 
B.  n = 1  II.  Slope of PV curve is $\frac{7}{5}\left(\frac{P}{V}\right)$  Q.  Work done by gas is less than process in B but more then that of D 
C.  $n=\frac{5}{3}$  III.  Slope of PV curve is zero  R.  Work done by gas$={P}_{1}{V}_{1}\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)$ 
D.  $n=\frac{7}{5}$  IV.  Slope of PV curve is $\frac{5}{3}\left(\frac{P}{V}\right)$  S.  Work done by gas is maximum 
Combination which is correct for adiabatic process is
Column I  Column II  Column III  
A.  n = 0  I.  Slope of PV curve is $\frac{P}{V}$  P.  Work done by gas this process is $\frac{5}{2}\left({P}_{1}{V}_{1}{P}_{2}{V}_{2}\right)$ 
B.  n = 1  II.  Slope of PV curve is $\frac{7}{5}\left(\frac{P}{V}\right)$  Q.  Work done by gas is less than process in B but more then that of D 
C.  $n=\frac{5}{3}$  III.  Slope of PV curve is zero  R.  Work done by gas$={P}_{1}{V}_{1}\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)$ 
D.  $n=\frac{7}{5}$  IV.  Slope of PV curve is $\frac{5}{3}\left(\frac{P}{V}\right)$  S.  Work done by gas is maximum 
Combination which is correct for a process in which temperature remains constant is
Column I  Column II  Column III  
A.  n = 0  I.  Slope of PV curve is $\frac{P}{V}$  P.  Work done by gas this process is $\frac{5}{2}\left({P}_{1}{V}_{1}{P}_{2}{V}_{2}\right)$ 
B.  n = 1  II.  Slope of PV curve is $\frac{7}{5}\left(\frac{P}{V}\right)$  Q.  Work done by gas is less than process in B but more then that of D 
C.  $n=\frac{5}{3}$  III.  Slope of PV curve is zero  R.  Work done by gas$={P}_{1}{V}_{1}\mathrm{ln}\left(\frac{{V}_{2}}{{V}_{1}}\right)$ 
D.  $n=\frac{7}{5}$  IV.  Slope of PV curve is $\frac{5}{3}\left(\frac{P}{V}\right)$  S.  Work done by gas is maximum 
Process for which pressure remains constant is
Column I  Column II  
P.  Process A → B  1.  Heat is gained 
Q.  Process B → C  2.  Heat is least 
R.  Process C → D  3.  Work done is zero 
S.  Process D → A  4.  Internal energy increases 
5.  Internal energy decreases 