Question

A vessel containing $1\;g$ of oxygen at a pressure of $10\;atm$ and a temperature of ${47^o}C$. 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 ${27^o}C$. Find the volume V of the vessel and the mass m of the oxygen that is leaked out in litres.
A) $m = \frac{1}{3}g,\;V = 0.082L$
B) $m = \frac{1}{3}g,\;V = 0.82L$
C) $m = \frac{2}{3}g,\;V = 0.082L$
D) $m = \frac{2}{3}g,\;V = 0.82L$

Answer 1

Hint: All the values in the system including pressure, mass and temperature are varying while the volume and the gas constant will be considered as constant. Thus, by using the ideal gas equation we can compare the values in both the situations to get the final result.

Formula used: Ideal gas equation $PV = nRT$

Complete Step by Step Solution:
We know that ideal gas equation is the medium to relate pressure, volume, number of moles and temperature to each other and is represented as follows:
$PV = nRT$
Where P is the pressure of the system, V is the volume of container, n is the number of moles of gas defined as the ratio of given mass to its molecular mass, T is the temperature and R is the gas constant.

Alternatively, we can write the above equation as follows:
$\frac{P}{{nT}} = \frac{R}{V}$
We know that, due to leak in the container, change in pressure will take place but the volume of the container will remain constant. Thus, we can say that for the given situation:
$\frac{{{P_1}}}{{{n_1}{T_1}}} = \frac{{{P_2}}}{{{n_2}{T_2}}}\;\;\;...(1)$

According to the question:
${P_1} = 10\;atm$
${P_2} = \frac{5}{8} \times 10 = \frac{{50}}{8}atm$
${{n}_{1}}=\frac{1}{32}\ \ \ \ \ \left[ \because \text{ number of moles =}\frac{given\ mass}{molecular\ mass} \right]$
${T_1} = {47^o}C + 273 = 320K$
${T_2} = {27^o}C + 273 = 300K$

Substituting given values in equation (1):
\[\frac{{10}}{{\frac{1}{{32}} \times 320}} = \frac{{\frac{{50}}{8}}}{{{n_2} \times 300}}\;\]
$ \Rightarrow {n_2} = \frac{{50 \times 320}}{{32 \times 10 \times 300 \times 8}} \Rightarrow \frac{1}{{48}}$
Here ${n_2}$ is the number of moles of oxygen left after the leakage. So the number of moles of gas leaked out of the vessel will be $n = {n_1} - {n_2} \Rightarrow \frac{1}{{32}} - \frac{1}{{48}} = \frac{1}{{96}}$.

Now, as per the definition of number of moles, we can calculate the mass of gas leaked out as follows:
${\rm{number of moles = }}\frac{{given\;mass}}{{molecular\;mass}}$
$ \Rightarrow \frac{1}{{96}} = \frac{m}{{32}}$
$ \Rightarrow m = \frac{{32}}{{96}} = \frac{1}{3}g$

As the volume of the container is constant, so it can be calculated by substituting the values in the equation of ideal gas as follows:
${P_1}V = {n_1}R{T_1}$
$ \Rightarrow V = \frac{{{n_1}R{T_1}}}{{{P_1}}} = \frac{{\frac{1}{{32}} \times 0.0821 \times 320}}{{10}}$
$ \Rightarrow V = 0.082L$
Hence, the volume V of the vessel and the mass m of the oxygen that is leaked out in litres is $0.082L$ and $\frac{1}{3}g$ respectively. Thus, option (A) is the correct answer.

Note: Please note that the value of R must be kept as per the value of pressure and temperature as the value of gas constant varies for different units. For the given question, the value of pressure was given in atm and the volume was to be calculated in litres, so the value of gas constant was kept as 0$0.0821\;{\rm{atmLmo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}$ while in terms of joules, the value is $8.314{\rm{ Jmo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}$.