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\[5\] moles of an ideal gas at ${27^o }C$ expands isothermally and reversibly from a volume of $6litres$ to $60litres$. The work done in $kJ$ is:
(A) $ - 14.7$
(B) $ - 28.72$
(C) $ + 28.72$
(D) $ + 14.7$

Last updated date: 20th Jun 2024
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Hint: A process can be said reversible when a change can be brought about in such a way that the process could be at any time reversed by an infinitesimal change. The number of moles, the volumes and temperature \[{27^ \circ }c\] means 300 kelvin which will be constant (isothermal) are given. The work done will be negative since its reversible process and expansion taking place at constant temperature.

Complete step-by-step solution:
There are two main types of processes-reversible and irreversible. Reversible process proceeds infinitely slowly by a series of equilibrium states such that the system and surroundings are always in near equilibrium with each other.
The work done in isothermal reversible process is given by the equation-
${W_{rev}} = - 2.303nRT\log \left( {\dfrac{{{V_f}}}{{{V_i}}}} \right)$ where
${V_f}$ is the final volume of the gas, ${V_i}$ is the initial volume, $T$ is the temperature, $R$ is the gas constant, $n$ is the number of moles of the gas.
$R$ is the universal gas constant that has the value $R = 8.314J{K^{ - 1}}mo{l^{ - 1}}$. The temperature is expressed in $kelvin$ and the volume expressed in $litres$ .
Now from the given data we can substitute the values in the above equation-
$n = 5$ , $T = 27 + 273K = 300K$ , $R = 8.314J{K^{ - 1}}mo{l^{ - 1}}$ , ${V_f} = 60litres$, ${V_i} = 6litres$
Substituting above values in the equation,
${W_{rev}} = - 2.303 \times 5 \times 8.314 \times 300\log \dfrac{{60}}{6} = - 28720J = - 28.72kJ$
So the work done is $ - 28.72kJ$ .

Additional information: Isothermal process is the process where the temperature remains constant.Expansion of gas in vacuum is called free expansion. No work is done during such expansion since the external pressure is zero.

The correct option is (B).

Note: The units generally used for expressing work is $kJ,cal$ . Work is positive when work is done on the system by the surroundings and work done is negative when work is done by the system on the surroundings. Maximum amount of work is done in an irreversible process.