Question

The entropy change involved in the isothermal reversible expansion of $2\mathrm{moles}$ of an ideal gas from a volume of $10{\mathrm{dm}}^3$ at $27°\mathrm{C}$ is to a volume of $100{\mathrm{dm}}^3$ at $27°\mathrm{C}$ is?

Answer 1

Step 1: Reversible isothermal expansion :

• It is defined as an infinitely slow increase in volume at a constant temperature.
• The mathematical expression for it is ${\mathrm{w}}_{\mathrm{rev}}=-\int \mathrm{P}∆\mathrm{V}=-{\mathrm{q}}_{\mathrm{rev}}$, where $\mathrm{P}$ = pressure, $\mathrm{V}$ = volume, ${\mathrm{q}}_{\mathrm{rev}}$ = reversible absorbing heat.
• Work is done is from the perspective of the system and q is heat flow.

Step 2: Given Information:

• Isothermal reversible expansion of $2\mathrm{moles}$ of an ideal gas from a volume of $10{\mathrm{dm}}^3$ at $27°\mathrm{C}$ is to a volume of $100{\mathrm{dm}}^3$.

Step 3: Calculation for entropy change:

Entropy change:

• “Entropy Change can be defined as the change in the state of disorder of a thermodynamic system that is associated with the conversion of heat or enthalpy into work.”
• The entropy change for reversible isothermal expansion can be given as:
  $\mathrm{\Delta S}=\mathrm{nR}\ln \frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}$
  $\mathrm{\Delta S}=(2.303)\left(2\right)(8.314)\log \frac{100}{10}$

$\mathrm{\Delta S}=38.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$

Therefore, the entropy change in process is $38.3\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}$.