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Pressure and Temperature

A closed tank...

Question

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does not allow the gas to leak across (Figure 2), the volume (in $m^{3})$ of the compartment A after the system attains equilibrium is.

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Solution

As in fig 2, the system attains equilibrium, so,
$P_{A} = P_{B}$ and $T_{A} = T_{B}$
$\frac{P_{A} V_{A}}{n_{A} T_{A} R} = \frac{P_{B} V_{B}}{n_{B} T_{B} R}$
$\frac{V_{A}}{V_{B}} = \frac{n_{A}}{n_{B}}$
$n_{A} = \frac{5}{400 R}$
$n_{B} = \frac{3}{300 R}$
Due to sliding of piston vol. of A will be increased by x and that of B will be decreased by x.
$V_{A} = 1 + x$
$V_{B} = 3 - x$
$\frac{1 + x}{3 - x} = \frac{\frac{5}{400 R}}{\frac{3}{300 R}}$
4(1+x)=5(3-x)
$4 x + 5 x = 11 \Rightarrow x = \frac{11}{9}$
Hence volume of container A will be
$V_{A} = 1 + \frac{11}{9} = \frac{20}{9} = 2.22$

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