Gujarati
12.Kinetic Theory of Gases
normal

Two containers $C_{1}$ and $C_{2}$ of volumes $V$ and $4 \,V$ respectively, hold the same ideal gas and are connected by a thin horizontal tube of negligible volume with a valve which is initially closed. The initial pressures of the gas in $C_{1}$ and $C_{2}$ are $p$ and $5 p$, respectively. Heat baths are employed to maintain the temperatures in the containers at $300 \,K$ and $400 \,K$, respectively. The valve is now opened. Select the correct statement.

A

The gas will flow from the hot container to the cold one and the process is reversible.

B

The gas will flow from one container to the other till the number of moles in two containers are equal.

C

A long time after the valve is opened, the pressure in both the containers will be $3 p$.

D

A long time af ter the valve is opened, number of moles of gas in the hot container will be thrice that of the cold one.

(KVPY-2019)

Solution

$(d)$ Let $n_{1}$ and $n_{2}$ are number of moles of gas present in container $C_{1}$ and $C_{2}$ respectively, before the value is opened.

Then, using $p V=n R T$.

$\text { and }$

$n_{1} =\frac{p V}{R(300)}$

$n_{2} =\frac{5 p(4 V)}{R(400)}$

When value is opened gas flows from $C_{2}$ to $C_{1}$ till pressure in $C_{1}$ and $C_{2}$ is equal. Let after equalisation of pressure in both $C_{1}$ and $C_{2}$, its value is $p_{0}$,

Then, using $p V=n R^{\prime} T$.

$p_{0} V=n_{1}^{\prime} R(300)\left(\right.$ Container $\left.C_{1}\right)$

and $p_{0} 4 V=n_{2}{ }^{\prime} R(400)$ (Container $C_{2}$ )

So, $\quad n_{1}^{\prime}=\frac{p_{0} V}{R(300)}$

and $\quad n_{2}^{\prime}=\frac{p_{0}(4 V)}{R(400)}$

As no gas is leaked from containers,

$n_{1}+n_{2}=n_{1}{ }^{\prime}: n_{2}{ }^{\prime}$

$\Rightarrow \quad \frac{p V}{R(300)}+\frac{20 p V}{R(400)}$ = $\quad \frac{p_{0} V}{R(300)}+\frac{4 p_{0} V}{R(400)}$

So, $p\left(\frac{1}{300}+\right.\left.\frac{20}{400}\right)$ $=p_{0}\left(\frac{1}{300}+\frac{4}{400}\right)$

$\Rightarrow \quad 4 p=p_{0}$

$\text { Now, } \frac{n_{2}{ }^{\prime}}{n_{1}{ }^{\prime}}=\frac{\frac{4 p_{0} V}{R(400)}}{\frac{p_{0} V}{R(300)}}=3$

Finally number of moles of $C_{2}$ is thrice of $C_{1}$.

Standard 11
Physics

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