Heat and Thermodynamics 4 Question 22
24. Two identical containers $A$ and $B$ with frictionless pistons contain the same ideal gas at the same temperature and the same volume $V$. The mass of the gas in $A$ is $m _A$ and that in $B$ is $m _B$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $2 V$. The changes in the pressure in $A$ and $B$ are found to be $\Delta p$ and $1.5 \Delta p$ respectively. Then
$(1998,2 M)$
(a) $4 m _A=9 m _B$
(b) $2 m _A=3 m _B$
(c) $3 m _A=2 m _B$
(d) $9 m _A=4 m _B$
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Solution:
- Process is isothermal. Therefore, $T=$ constant. $\left(p \propto \frac{1}{V}\right)$ volume is increasing, therefore, pressure will decrease. In chamber $\boldsymbol{A} \rightarrow$
$$ \begin{aligned} -\Delta p=\left(p _A\right) _i-\left(p _A\right) _f & =\frac{n _A R T}{V}-\frac{n _A R T}{2 V} \\ & =\frac{n _A R T}{2 V} \end{aligned} $$
In chamber $\boldsymbol{B} \rightarrow$
$$ \begin{aligned} -1.5 \Delta p=\left(p _B\right) _i-\left(p _B\right) _f & =\frac{n _B R T}{V}-\frac{n _B R T}{2 V} \\ & =\frac{n _B R T}{2 V} \end{aligned} $$
From Eqs. (i) and (ii)
$$ \text { or } \quad \begin{array}{ll} \frac{n _A}{n _B} & =\frac{1}{1.5}=\frac{2}{3} \quad \text { or } \quad \frac{m _A / M}{m _B / M}=\frac{2}{3} \\ \frac{m _A}{m _B} & =\frac{2}{3} \text { or } 3 m _A=2 m _B \end{array} $$
