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Question: Three moles of an ideal gas are expanded isothermally from a volume of \[300\,c{m^3}\]to\[2.5\,L\] a...

Three moles of an ideal gas are expanded isothermally from a volume of 300cm3300\,c{m^3}to2.5L2.5\,L at 300K{\text{300}}\,{\text{K}} against a pressure of 1.9atm{\text{1}}{\text{.9}}\,{\text{atm}}. The work done in joules is:
A)  - 423.56J{\text{ - 423}}{\text{.56}}\,{\text{J}}
B)  + 423.56J{\text{ + 423}}{\text{.56}}\,{\text{J}}
C)  - 4.18J{\text{ - 4}}{\text{.18}}\,{\text{J}}
D)  + 4.8J{\text{ + 4}}{\text{.8}}\,{\text{J}}

Explanation

Solution

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 V1{{\text{V}}_1} to V2{{\text{V}}_2} at constant temperature and pressure.

Formula used: When ideal gas is expanded isothermally from volume V1{{\text{V}}_1} to V2{{\text{V}}_2} the work done is given as follows:
w = - pdv\,{\text{w = - pdv}}
w = - p(V2V1)\,{\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 dv{\text{dv}}, initial volume is V1{{\text{V}}_1}, and the final volume is V2{{\text{V}}_2}.

Complete step-by-step answer:
First, we have to convert the initial volume from centimeter to liters as follows:
1L = 103cm3{\text{1}}\,{\text{L = 1}}{{\text{0}}^3}\,{\text{c}}{{\text{m}}^{\text{3}}}
300cm3=300cm3×1L103cm3{\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}}}}}
300cm3=0.3L{\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.
w = - p(V2V1)\,{\text{w = - p}}\left( {{V_2} - {V_1}} \right)
Here, 1.9atm{\text{1}}{\text{.9}}\,{\text{atm}} for P, 0.3L0.3\,\,{\text{L}} for V1{{\text{V}}_1}, and 2.5L{\text{2}}{\text{.5}}\,{\text{L}} for V2{{\text{V}}_2}.
w = - p(V2V1)\,{\text{w = - p}}\left( {{V_2} - {V_1}} \right)
w = - (1.9atm×(2.5L0.3L))\,{\text{w = - }}\left( {{\text{1}}{\text{.9}}\,{\text{atm}} \times \left( {2.5\,L - 0.3\,L} \right)} \right)
w=4.18atmL\,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.
1atmL=101.325J\,1\,atm\,L = 101.325\,J
4.18atmL=4.18atmL×101.325J1atmL\, - 4.18\,atm\,L = \,\dfrac{{ - 4.18\,atm\,L \times 101.325\,J}}{{1\,atm\,L}}
423.5385J- 423.5385\,J
Thus, work done in units of joules is 423.5385J - 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.