Question

Calculate the final pressure after removing partition between the two chambers having two different gases of same volume at same temperature and different pressure.
[Assumption: Temperatures are equal before and after removing the partition.]


Given, pressure $P_1=1\text{atm}$, pressure $P_2=2\text{atm}$

• A. $1.5\text{atm}$
• B. $1.25\text{atm}$
• C. $1.34\text{atm}$
• D. $1.75\text{atm}$

Answer 1

The correct option is A $1.5\text{atm}$

Given,
$P_1=1\text{atm};P_2=2\text{atm}{V}_1=V_2=V;T_1=T_2=T$

At constant temperature and volume,

From, $PV=nRT....\left(1\right)$

$\Rightarrow P=\frac{nRT}{V}[\because T\text{and}V=\text{cons}]$

$\Rightarrow P\propto n$

Let $n_1$ and $n_2$ be the no. of moles of gas in left and right chamber respectively,

$n_1=\frac{P_1{V}_1}{R{T}_1}=\frac{1\times V}{RT}=\frac{V}{RT}$

$n_2=\frac{P_2{V}_2}{R{T}_2}=\frac{2V}{RT}$

Finally, after removing partition, the given system becomes a single chamber,

$n_{total}=n_1+n_2=\frac{V}{RT}+\frac{2V}{RT}=\frac{3V}{RT}$

Hence, the final volume will be,

$V_f=V_1+V_2=2V$

From (1) we get,

$P_f{V}_f=n_{total}RT$

$\Rightarrow P_f=\frac{\frac{3V}{RT}\times RT}{2V}$

$\Rightarrow P_f=1.5\text{atm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.

Why this question? Key concept: At constant temperature and volume, pressure of a gas $\left(P\right)$ is $P\propto n$ Where, $n=$ No. of moles in the gas