Entropy Change - Practice Questions & MCQ

Mathematical Definition of Entropy

For a reversible isothermal process, Clausius defined it as the integral of all the terms involving heat exchange (q) divided by the absolute temperature T.

$\mathrm{dS}=\frac{\mathrm{dq}_{\mathrm{rev}}}{\mathrm{T}}$ or $\Delta \mathrm{S}=\frac{\mathrm{q}_{\mathrm{rev}}}{\mathrm{T}}$

Unit of entropy is $\frac{\mathrm{J}}{\mathrm{mol}-\mathrm{K}}$ 

Here mol^-1 is also used as entropy being an extensive property depends upon the amount of the substance.

 

Entropy Changes in different processes:

1. Isothermal reversible process

For a reversible isothermal process, $\Delta \mathrm{E}=0$

So $\mathrm{q}=-\mathrm{w}$

$\therefore \Delta \mathrm{S}=\frac{-\mathrm{w}}{\mathrm{T}}=\frac{2.303 \mathrm{nRT} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)}{\mathrm{T}}$

$\therefore \Delta \mathrm{S}=2.303 \mathrm{nR} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)=2.303 \mathrm{nR} \log \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)$

 

2. Adiabatic reversible process

As $\mathrm{q}=0$, so $\Delta \mathrm{S}=0$

Note: Reversible adiabatic process is also called as Isentropic process

 

3. Isobaric process:

$\Delta \mathrm{S}=2.303 \mathrm{nC}_{\mathrm{P}} \log \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)=2.303 \mathrm{nC}_{\mathrm{P}} \log \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)$

 

4. Isochoric process:

$\Delta \mathrm{S}=2.303 \mathrm{nC}_{\mathrm{V}} \log \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)=2.303 \mathrm{nC}_{\mathrm{V}} \log \left(\frac{\mathrm{P}_2}{\mathrm{P}_1}\right)$

 

5. Entropy change in a process where both the Temperature as well as Volume or Pressure is changing

$\Delta \mathrm{S}=\int \frac{\mathrm{dq}}{\mathrm{T}}=\int \frac{(\mathrm{dE}-\mathrm{dw})}{\mathrm{T}}$

$\Delta S=\int \frac{\mathrm{nC}_{\mathrm{v}} \mathrm{dT}+\mathrm{PdV}}{\mathrm{T}}=\int_{\mathrm{T}_1}^{\mathrm{T}_2} \frac{\left(\mathrm{nC}_{\mathrm{v}} \mathrm{dT}\right)}{\mathrm{T}}+\int_{\mathrm{V}_1}^{\mathrm{V}_2} \frac{(\mathrm{nRdV})}{\mathrm{V}}$

$\Delta \mathrm{S}=\mathrm{nC}_{\mathrm{v}} \ln \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)+\mathrm{nR} \ln \left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)$

The above equation can also be written in terms of Pressure as 

$\Delta \mathrm{S}=\mathrm{nC}_{\mathrm{p}} \ln \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)+\mathrm{nR} \ln \left(\frac{\mathrm{P}_1}{\mathrm{P}_2}\right)$

Note: Remember the above general formula for the change in entropy.

 

6. Entropy change in irreversible processes:

Suppose a system at higher temperature T₁ and its surroundings is at lower temperature T₂.  'q' amount of heat goes irreversibly from the system to the surroundings.

$\Delta \mathrm{S}_{\text {system }}=-\frac{\mathrm{q}}{\mathrm{T}_1}$

$\Delta \mathrm{S}_{\text {surroundings }}=+\frac{\mathrm{q}}{\mathrm{T}_2}$

$\Delta \mathrm{S}_{\text {process }}=\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surroundings }}=-\frac{\mathrm{q}}{\mathrm{T}_1}+\frac{\mathrm{q}}{\mathrm{T}_2}=\mathrm{q} \frac{\left[\mathrm{T}_1-\mathrm{T}_2\right]}{\mathrm{T}_1 \mathrm{~T}_2}$

$\begin{aligned} & \because \mathrm{T}_1>\mathrm{T}_2 \\ & \therefore \mathrm{~T}_1-\mathrm{T}_2>0\end{aligned}$

$\therefore \Delta \mathrm{S}_{\text {process }}>0$

So entropy increases in an irreversible process like conduction, radiation, etc.

 

7. Entropy changes during phase transition:

$\Delta \mathrm{S}=\mathrm{S}_2-\mathrm{S}_1=\frac{\mathrm{q}_{\mathrm{rev}}}{\mathrm{T}}=\frac{\Delta \mathrm{H}}{\mathrm{T}}$

 

8. Entropy change when liquid is heated:

When a definite amount of liquid of mass 'm' and specific heat 's' is heated

Let us suppose a small amount of heat dq is added and as a result the temperature of the body increases by dT temperature

$\mathrm{dq}=\mathrm{m} \times \mathrm{s} \times \mathrm{dT}$

$\therefore \mathrm{dS}=\frac{\mathrm{dq}}{\mathrm{T}}=\frac{\mathrm{m} \times \mathrm{s} \times \mathrm{dT}}{\mathrm{T}}$

$\therefore \Delta \mathrm{S}=\mathrm{m} \times \mathrm{s} \times \log \frac{\mathrm{T}_2}{\mathrm{~T}_1}$

 

9. Entropy Change in Mixing of Ideal Gases:

Suppose n₁ mole of gas 'P' and n₂ mole of gas Q' are mixed; then total entropy change can be calculated as:

$\Delta \mathrm{S}=-2.303 \mathrm{R}\left[\mathrm{n}_1 \log _{10} \mathrm{X}_1+\mathrm{n}_2 \log _{10} \mathrm{X}_2\right]$

Here X₁ and X₂ are mole fractions of gases P and Q respectively.

 

$\Delta \mathrm{S} / \mathrm{mol}=-2.303 \mathrm{R} \frac{\left[\mathrm{n}_1 \log _{10} \mathrm{X}_1\right.}{\mathrm{n}_1+\mathrm{n}_2}+\frac{\left.\mathrm{n}_2 \log _{10} \mathrm{X}_2\right]}{\mathrm{n}_1+\mathrm{n}_2}$

$\Delta \mathrm{S} / \mathrm{mol}=-2.303 \mathrm{R}\left[\mathrm{X}_1 \log _{10} \mathrm{X}_1+\mathrm{X}_2 \log _{10} \mathrm{X}_2\right]$

It can be seen that the above expression is always positive for $\Delta \mathrm{S}$.