Question

A vessel of volume $30m^{3}$ contains an ideal gas at room temperature and pressure $5.16bar$ . The gas is allowed to leak till its pressure falls to atmospheric pressure. Assuming that the process to be isothermal, the number of moles of the gas that have leaked are $\times 10^{3}$
(Consider $R=8.32Jmol^{- 1 }K^{- 1}$ , atmospheric pressure $=1bar=10^{5}Pa$ )

Answer 1

Before leakage,
$PV=n_{1}RT\ldots ..\left(\right.i\left.\right)$
After leakage,
$P^{'}V=n_{2}RT\ldots ..\left(\right.ii\left.\right)$
Number of moles of gas leaked is given by,
$\left(n_{1} - n_{2}\right)$
Subtracting equation (ii) from equation (i),
$n_{1}-n_{2}=\frac{PV}{RT}-\frac{P^{'} V}{RT}$
$\therefore n_{1}-n_{2}=\frac{V}{RT}\left(P - P^{'}\right)$
$=\frac{30}{8 . 32 \times \left(\right. 27 + 273 \left.\right)}\times \left(\right.5.16-1\left.\right)\times \left(10\right)^{5}$
$=\frac{30 \times 4 . 16 \times 10^{5}}{8 . 32 \times 300}$
$\therefore n_{1}-n_{2}=5\times 10^{3}$