Question

Three moles of an ideal gas are expanded isothermally from a volume of \[300\,c{m^3}\]to\[2.5\,L\] at \[{\text{300}}\,{\text{K}}\] against a pressure of \[{\text{1}}{\text{.9}}\,{\text{atm}}\]. The work done in joules is:
A) \[{\text{ - 423}}{\text{.56}}\,{\text{J}}\]
B) \[{\text{ + 423}}{\text{.56}}\,{\text{J}}\]
C) \[{\text{ - 4}}{\text{.18}}\,{\text{J}}\]
D) \[{\text{ + 4}}{\text{.8}}\,{\text{J}}\]

Answer 1

Hint: In general work is defined as the product of force and displacement. Work is a path function whose values depend on the following. Here, isothermal expansion of the gas is given indicates the gas is expanded from volume \[{{\text{V}}_1}\] to \[{{\text{V}}_2}\] at constant temperature and pressure.

Formula used: When ideal gas is expanded isothermally from volume \[{{\text{V}}_1}\] to \[{{\text{V}}_2}\] the work done is given as follows:
\[\,{\text{w = - pdv}}\]
\[\,{\text{w = - p}}\left( {{V_2} - {V_1}} \right)\]
Here, work is represented as w, the pressure is P, change in volume is represented as \[{\text{dv}}\], initial volume is \[{{\text{V}}_1}\], and the final volume is \[{{\text{V}}_2}\].

Complete step-by-step answer:
First, we have to convert the initial volume from centimeter to liters as follows:
\[{\text{1}}\,{\text{L = 1}}{{\text{0}}^3}\,{\text{c}}{{\text{m}}^{\text{3}}}\]
\[{\text{300}}\,{\text{c}}{{\text{m}}^{\text{3}}} = \dfrac{{{\text{300}}\,{\text{c}}{{\text{m}}^{\text{3}}}{{ \times 1}}\,{\text{L}}}}{{{\text{1}}{{\text{0}}^3}\,{\text{c}}{{\text{m}}^{\text{3}}}}}\]
\[{\text{300}}\,{\text{c}}{{\text{m}}^{\text{3}}} = 0.{\text{3}}\,{\text{L}}\]
Here, to find out the work we have to use the above formula.
\[\,{\text{w = - p}}\left( {{V_2} - {V_1}} \right)\]
Here, \[{\text{1}}{\text{.9}}\,{\text{atm}}\] for P, \[0.3\,\,{\text{L}}\] for \[{{\text{V}}_1}\], and \[{\text{2}}{\text{.5}}\,{\text{L}}\] for \[{{\text{V}}_2}\].
\[\,{\text{w = - p}}\left( {{V_2} - {V_1}} \right)\]
\[\,{\text{w = - }}\left( {{\text{1}}{\text{.9}}\,{\text{atm}} \times \left( {2.5\,L - 0.3\,L} \right)} \right)\]
\[\,w = - 4.18\,atm\,L\]
Here, work done is obtained in the unit of atmosphere liters. To convert work in joules following conversion factor is used.
\[\,1\,atm\,L = 101.325\,J\]
\[\, - 4.18\,atm\,L = \,\dfrac{{ - 4.18\,atm\,L \times 101.325\,J}}{{1\,atm\,L}}\]
\[ - 423.5385\,J\]
Thus, work done in units of joules is \[ - 423.5385\,J\].

Here, option (A) is the correct answer for the given question.

Note: Here, in the formula of the work negative sign is used. The significance of the sign indicates whether work is done on the system or work is done by the system. During the expansion of the gas change in volume is positive therefore, work done is negative indicates work done by the system on the surrounding. During compression of the gas change in volume is negative therefore, work done positively indicates work done by the system by surrounding.