Question

Calculate the work done when $1.6 \times {10^{ - 2}}kg$ of oxygen at $300K$ are expanded isothermally and reversibly till the volume is increased by four times.

Answer 1

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. Here we also need to calculate the number of moles of oxygen. It can be calculated with the help of molar mass of oxygen.

Complete step by step answer:
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-
${W_{rev}} = - 2.303nRTlog \dfrac{{{V_f}}}{{{V_i}}}$ 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.
Now first we need to find out the number of moles of oxygen gas. We know,
Number of moles $ = \dfrac{{given \, mass}}{{molar\, mass}}$
One mole of oxygen $ = 32gmo{l^{ - 1}}$
Number of moles in $1.6 times {10^{ - 2}}kg,or,16g$ of oxygen $ = \dfrac{{16}}{{32}} = 0.5moles$ of oxygen.
It is given that volume is increased four times. So the final volume is four times the initial volume. [{V_f} = 4{V_i}] .
Now substituting the values in above equation-
$n = 0.5moles$
$R = 8.314J/K/mol$
$T = 300K$
$V_f = 4V_i$
${W_{rev}} = - 2.303nRTlog \dfrac{{{V_f}}}{{{V_i}}}$
Substituting the values-
${W_{rev}} = - 2.303 \times 0.5 \times 300 \times 8.314log \dfrac{{4{V_i}}}{{{V_i}}} = - 2.303 \times 150 \times 8.314log 4 = - 1729.16J = - 1.729kJ$
Hence the work done is $ - 1.729kJ$ .

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.

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.