Question

Oxygen is filled in a closed metal jar of volume 1.0 × 10−3 m3 at a pressure of 1.5 × 105 Pa and temperature 400 K. The jar has a small leak in it. The atmospheric pressure is 1.0 × 105 Pa and the atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding.

Answer 1

The mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalise with the surrounding is $=0.16 \mathrm{g}$

Explanation:

Step 1:

$\mathrm{V}_{1}=1.0 \times 10^{-3} \mathrm{m}^{3}$

$T_{1}=400 K$

$P_{1}=1.5 \times 10^{5} \mathrm{Pa}$

$\mathrm{P}_{2}=1.0 \times 10^{5} \mathrm{Pa}$

$T_{2}=300 M=32 g$

Amount of moles in the front bowl

$n_{1}=\frac{P_{1} V_{1}}{R T_{1}}$

Step 2:

Gas volume is given when the external pressure is equal to

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

$V_{2}=\frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}}$

$V_{2}=\frac{1.5 \times 10^{5} \times 1.0 \times 10^{-3} \times 300}{1.0 \times 10^{5} \times 400}$

$V_{2}=\frac{15 \times 10^{2} \times 30}{10^{5} \times 400}$

$V_{2}=\frac{450 \times 10^{2}}{10^{5} \times 400}$

$V_{2}=\frac{45 \times 10^{-5} \times 10^{2}}{40}$

$V_{2}=\frac{45 \times 10^{-3}}{40}$

$V_{2}=\frac{9 \times 10^{-3}}{8}$

$V_{2}=1.125 \times 10^{-3}$

Step 3:

Net amount of leaked gas = $\mathrm{V}_{2}-\mathrm{V}_{1}$

$=1.125 \times 10^{-3}-1.0 \times 10^{-3}$

$=1.25 \times 10^{-4} \mathrm{m}^{3}$

Let n2 be the number of yielded gas moles. When we apply state equation on this amount of gas, we get

$n_{2}=\frac{P_{2} V_{2}}{R T_{2}}$

$n_{2}=\frac{1.0 \times 10^{5} \times 1.25 \times 10^{-4}}{8.3 \times 300}$

$n_{2}=\frac{10 \times 1.25}{2490}$

$n_{2}=\frac{12.5}{2490}$

$=0.005$

leaked gas mass= $32 \times 0.005=0.16 \mathrm{g}$