Question

Perfect gas of ${\text{280}}\,{\text{mmol}}$ occupies ${\text{12}}{\text{.7}}\,{\text{L}}$ at ${\text{310}}\,{\text{K}}$. Calculate the work done when the gas expands
A) isothermally against a constant external pressure of ${\text{0}}{\text{.25}}\,{\text{atm}}$
B) isothermally and reversibly
C) into vacuum until its volume has increased by ${\text{3}}{\text{.3}}\,{\text{L}}$

Answer 1

Work is one of the thermodynamic properties which is a product of force and displacement.
Work is the path function that depends on the path followed by the system.
Work is done on the system by the surrounding it is positive and when it is done by the system on the surrounding is then as negative.

Complete step by step solution:
Here, the moles of has given are ${\text{280}}\,{\text{mmol}}$ that is ${\text{280}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{mol}}$. The initial volume of gas is ${\text{12}}{\text{.7}}\,{\text{L}}$, and the temperature is ${\text{310}}\,{\text{K}}$ and the pressure is ${\text{0}}{\text{.25}}\,{\text{atm}}$.
Here, the volume is increased by ${\text{3}}{\text{.3}}\,{\text{L}}$, hence, the final volume is $\left( {{\text{12}}{\text{.7}}\,{\text{ + 3}}{\text{.3}}} \right)\,{\text{L}} = 16\,{\text{L}}$.
A) gas is expanded isothermally against a constant external pressure of ${\text{0}}{\text{.25}}\,{\text{atm}}$.
It is calculated using the formula as follows:
${\text{w = }} - {\text{pdV}}$
Here, first, determine the volume change.
${\text{dV = }}{{\text{V}}_2} - {{\text{V}}_1}$
Substitute ${\text{12}}{\text{.7}}\,{\text{L}}$ for ${{\text{V}}_1}$ and $16\,{\text{L}}$ for ${{\text{V}}_2}$.
${\text{dV = }}{{\text{V}}_2} - {{\text{V}}_1}$
${\text{dV = }}16\,{\text{L}} - {\text{12}}{\text{.7}}\,{\text{L}}$
${\text{dV = 3}}{\text{.3}}\,\,{\text{L}}$
Substitute ${\text{0}}{\text{.25}}\,{\text{atm}}$ for P, ${\text{3}}{\text{.3}}\,\,{\text{L}}$ for ${\text{dV}}$.
${\text{w = }} - {\text{pdV}}$
${\text{w = }} - \left( {{\text{0}}{\text{.25}}\,{\text{atm}}} \right) \times \left( {{\text{3}}{\text{.3}}\,\,{\text{L}}} \right)$
${\text{w = }} - 0.825\,{\text{atm}}\,{\text{L}}$
Convert the atm L to joules as follows:
$1\,{\text{atm}}\,{\text{L}} = 101.3\,{\text{J}}$
$- 0.825\,{\text{atm}}\,{\text{L}} = \dfrac{{ - 0.825\,{\text{atm}}\,{\text{L}}}}{{1\,{\text{atm}}\,{\text{L}}}} \times 101.3\,{\text{J}}$
$- 0.825\,{\text{atm}}\,{\text{L}} = - 83.5725\,{\text{J}}$
Thus, the work done is obtained when gas is expanded isothermally against a constant external pressure of ${\text{0}}{\text{.25}}\,{\text{atm}}$ is $- 83.6\,{\text{J}}$.
B) gas expand isothermally and reversibly
Work done is obtained by using the following formula.
${\text{w = }} - {\text{nRT}}\,{\text{ln}}\dfrac{{{{\text{V}}_{\text{1}}}}}{{{{\text{V}}_{\text{2}}}}}$
Here, substitute ${\text{280}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{mol}}$ for n, $8.314\,{\text{J}}\,{{\text{K}}^{{\text{ - 1}}}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ for R, ${\text{310}}\,{\text{K}}$ for T, ${\text{12}}{\text{.7}}\,{\text{L}}$ for ${{\text{V}}_1}$, and $16\,{\text{L}}$ for ${{\text{V}}_2}$.
${\text{w = }} - \left( {{\text{280}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{mol}} \times 8.314\,{\text{J}}\,{{\text{K}}^{{\text{ - 1}}}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}} \times {\text{310}}\,{\text{K}} \times \,{\text{ln}}\dfrac{{{\text{12}}{\text{.7}}\,{\text{L}}}}{{16\,{\text{L}}}}} \right)$
${\text{w = }} - \left( {{\text{280}} \times {\text{1}}{{\text{0}}^{ - 3}}\,{\text{mol}} \times 8.314\,{\text{J}}\,{{\text{K}}^{{\text{ - 1}}}}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}} \times {\text{310}}\,{\text{K}} \times \,\left( { - 0.230987} \right)} \right)$
${\text{w = }} - 166.69\,{\text{J}}$
Thus, the work done when a gas expands isothermally and reversibly is $- 166.7\,{\text{J}}$.
C) gas expand into vacuum
When gas is expanded into the vacuum there is no external pressure that is external pressure is zero.
${\text{w = }} - {\text{pdV}}$
Here, p is zero,
${\text{w = }}0$
Thus, work done in case of the free expansion is zero.

Note: In the case of the expansion of gas final volume is higher than the initial volume, the work obtained is negative indicates work done by the system on the surrounding.In case of the compression of the gas initial volume is higher than the final volume, hence work obtained is positive indicates work done on the system by the surrounding.