Question

In a thermally isolated system, two boxes filled with an ideal gas are connected by a valve. When the valve is in closed position, states of the box $1$ and $2$ respectively, are ($1$ atm, $V, T)$ and $(0.5$ atm, $4 \,V, T)$. When the valve is opened, then the final pressure of the system is approximately

• A. 0.5 atm
• B. 0.6 atm
• C. 0.75 atm
• D. 1.0 atm

Answer 1

Answer: (B)

Given situation is

Let final temperature after opening the valve is $T_{f}$,
then $\Delta W_{\text {ext }}-0$ and $\Delta Q_{\text {ext }}-0$
So, from first law of thermodynamics,
$\Delta U =0 $
$\Rightarrow n_{1} C_{V} T+n_{2} C_{V} T =\left(n_{1}+n_{2}\right) C_{V} T_{f} $
$\Rightarrow T_{f} =T $
Now, by gas equation, we have
As, $ n_{1}+n_{2}=n$
$\Rightarrow \frac{V}{R T}+\frac{4 V \times 0.5}{R T}=\frac{5 V \times p_{1}}{R T}$
$\Rightarrow p=0.6 \,atm$