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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 ............... $atm$
$0.5$
$0.6$
$0.75$
$1.0$
Solution
(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
$\text { As, } n_1+n_2=n$
$\Rightarrow \frac{V}{R T}+\frac{4 V \times 0.5}{R T} =\frac{5 V \times p_1}{R T}$
$\Rightarrow p =0.6 \,atm$
