Two moles of an ideal gas occupy a volume V.The gas expands isothermally and reversibly to a volume 3V . Calculate the change in entropy of the gas.

Entropy can be defined as the measure of a system’s thermal energy per unit temperature that is unavailable for doing useful work. It is also the measure of randomness of system.

Defining first law of thermodynamics

The first law of thermodynamics states that the net heat energy supplied to the system is equal to sum of change in internal energy and work done by the system.

Defining the isothermal process

It is a type of thermodynamic process in which the temperature of the system remain constant.

Formula of change in entropy

$∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$

Where, $∆\mathrm{S}$ is the change in entropy of gas, _$\mathrm{Q}$is the heat gain by the gas and $\mathrm{T}$ is the temperature of gas.

Then mathematically it can be written as

$\mathrm{Q}=\mathrm{W}+∆\mathrm{U}$

Where, $\mathrm{Q}$is the heat energy supplied to the system, $\mathrm{W}$ is the work done by the system and _{$∆\mathrm{U}$} is change in internal energy of the system.

Work done and change in internal energy in an isothermal process

_{$\mathrm{W}=\mathrm{nRTIn}\frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}$}

Where, $\mathrm{W}$ is the work done by the system, n the number of moles, is the temperature, ${\mathrm{V}}_{\mathrm{i}}$ is the initial volume, ${\mathrm{V}}_{\mathrm{f}}$ is the final volume, and $\mathrm{R}$ is the universal gas constant which is equal to $8.314\mathrm{J}/\mathrm{K}$_.

The change in internal energy of the system, $∆\mathrm{U}=0$

Here, change in internal energy of system is zero because the temperature is constant.

Consider the given data as below.

Initial volume of gas is ${\mathrm{V}}_{\mathrm{i}}=\mathrm{V}$

Final volume of gas is ${\mathrm{V}}_{\mathrm{f}}=3\mathrm{V}$

Number of moles is $\mathrm{n}=2$.

Using formula

_{$∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$}

From first law of thermodynamics

_{$∆\mathrm{S}=\frac{\mathrm{W}+∆\mathrm{U}}{\mathrm{T}}$}

Here, for isothermal system can be written as

_{$∆\mathrm{S}=\frac{\mathrm{nRT}\mathrm{In}\left({\displaystyle \frac{{\mathrm{V}}_{\mathrm{f}}}{{\mathrm{V}}_{\mathrm{i}}}}\right)}{\mathrm{T}}$}

Now, putting the values of constants in above relation

_{$$\begin{gathered} ∆\mathrm{S}=\frac{\mathrm{nRT}\mathrm{In}\left({\displaystyle \frac{3\mathrm{V}}{\mathrm{V}}}\right)}{\mathrm{T}} \\ =2\times 8.314\times \mathrm{In}3 \\ =18.10\mathrm{J}/\mathrm{K} \end{gathered}$$}

Thus, the change in entropy of gas is _{$18.10\mathrm{J}/\mathrm{K}$.}