Question

The volume of containers A and B, connected by a tube and a closed valve are $V$ and $4V$ respectively. Both the containers A and B have the same ideal gas, at pressures $5.0\times {10}^5\text{Pa}$ and $1.0\times {10}^5\text{Pa}$ and temperatures $300\text{K}$ and $400\text{K}$ respectively. The valve is opened to allow the pressure to equalise, but the temperature of each container is kept constant at its initial value. Find the common pressure in the containers.

• A. $2.5\times {10}^5\text{Pa}$
• B. $2.0\times {10}^5\text{Pa}$
• C. $3.0\times {10}^5\text{Pa}$
• D. $1.5\times {10}^5\text{Pa}$

Answer 1

The correct option is B $2.0\times {10}^5\text{Pa}$


Given,
Pressure of gas in container A $\left(P_A\right)=5.0\times {10}^5\text{Pa}$
Pressure of gas in container B $\left(P_B\right)=1.0\times {10}^5\text{Pa}$
Temperature of gas in container A $\left(T_A\right)=300\text{K}$
Temperature of gas in container B $\left(T_B\right)=400\text{K}$
From the figure, volume of container A and B are $V$ and $4V$.
$\because P_A>P_B$, to equalise the pressure, there will be movement of gas molecules from container A to container B.

Let the common pressure be $P$, $n_A$ and $n_B$ be the number of moles of the gas in containers A and B before opening the valve and $n_A^{\prime }$ and $n_B^{\prime }$ be the number of moles of the gas in containers A and B after opening the valve.
Since the quantity of gas in both the containers combined remains constant, we can say that,
$n_A+n_B=n_A^{\prime }+n_B^{\prime }...\left(1\right)$
By using, ideal gas equation, we can write $\left(1\right)$ as
$\frac{P_A{V}_A}{T_A}+\frac{P_B{V}_B}{T_B}=\frac{P{V}_A}{T_A}+\frac{P{V}_B}{T_B}$
$\frac{5\times {10}^5\times V}{300}+\frac{1\times {10}^5\times 4V}{400}=\frac{P\times V}{300}+\frac{P\times 4V}{400}$
$\Rightarrow (\frac{5}{3}+1)\times {10}^5=P\times \frac{4}{3}$
$\Rightarrow P=2\times {10}^5\text{Pa}$
Thus, option (b) is the correct answer.