Two chemically inactive gases are contained separately in two bulbs of same capacity at pressure P1 and P2 respectively if 2 bulbs are connected total pressure of gaseous mixture would be:
a. equal to (P1+P2)
b. greater than (P1+P2)
​c. less than (P1+P2)
​d. equal to (P1xP2)

​I want answer with full solution as soon as possible.

Dear Student,

Please find below the solution to the asked query.

The correct answer is option c) less than (P₁+P₂)

Let, volume of each bulb = V

So, total volume after mixing = V + V = 2V

P₁ and P₂ are the individual pressures of the respective gases, and not the partial pressure in the mixture. 

So, according to Boyle's law:

P₁V₁ = P₂V₂

Now, for first gas in first bulb, 

P₁V = p₁(2V)
where, p₁ = partial pressure of gas in the mixture
=>  $$\begin{gathered} p_1 = \frac{P_{1}V}{2V} \\ \Rightarrow p_1 = \frac{P_1}{2} \end{gathered}$$

Similarly, for the second gas in second bulb:

P₂V = p₂ (2V)
where, p₂ = partial pressure of second gas in the mixture.
$\Rightarrow p_2 = \frac{P_2}{2}$

Nw, according to Dalton's law, the pressure exerted by a mixture of two or more non-reactive gases in a mixture is equal to the sum of partial pressure of the individual gases.

So, p_mix = p₁ + p₂
=> p_mix = $\frac{P_1}{2}+\frac{P_2}{2}$
=> p_mix = $\frac{1}{2}(P_1+P_2)$

Thus, p_mix < (P₁ + P₂)

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Regards
Akanksha Jain