Question

A vessel containing $1g$ of oxygen at a pressure of $10atm$ and a temperature of ${47}^{\circ }C$. It is found that because of a leak, the pressure drops to $\frac{5}{8}th$ of its original value and the temperature falls to ${27}^{\circ }C$. Find the volume V of the vessel and the mass m of oxygen that is leaked out in litres.

• A. m=$\frac{1}{3}$ g, V=0.082 L
• B. m=$\frac{1}{3}$ g, V=0.82 L
• C. m=$\frac{2}{3}$ g, V=0.082 L
• D. m=$\frac{2}{3}$ g, V=0.82 L

Answer 1

The correct option is A m=$\frac{1}{3}$ g, V=0.082 L

The pressure, temperature and the number of moles of oxygen in the vessel change due to leak, while the volume remains fixed. Hence using $PV=nRT$, we have
$\left(\frac{P_1}{n_1{T}_1}\right)=\left(\frac{P_2}{n_2{T}_2}\right)⟹n_2=\frac{320}{300}\times \frac{5}{8}\times \frac{1}{32}=\frac{1}{48}moles$ (1 g of oxygen has $\frac{1}{32}$ moles)
Mass leaking out$=1-n_2\left(32\right)=1-\left(\frac{2}{3}\right)=\frac{1}{3}g$
Volume of vessel$=\frac{n_{1}R{T}_1}{P_1}$
$⟹V=\frac{1\times 0.082\times 320}{32\times 10}=0.082L$