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Entropy Change MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Syllabus

Quick Facts

Calculation Of Changes In S For Different Process is considered one the most difficult concept.
9 Questions around this concept.

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EASY

When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature, the change in entropy of gas ( $\Delta S$ ) is :

$\mathrm{C_{p,m} \ ln 2}$

$\mathrm{C_{v,m}\ ln 2}$

$\mathrm{ R \ ln 2}$

$\mathrm{(C_{v,m} - R ) ln 2}$

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For a sample of perfect gas when its pressure is changed isothermally from $P_i$ to $P_f$, the entropy change is given by:

$\Delta S=n R \ln \frac{P_f}{P_i}$

$\Delta S=n R \ln \frac{P_i}{P_f}$

$\Delta S=n R T \ln \frac{P_f}{P_i}$

$\Delta S=R T \ln \frac{P_i}{P_f}$

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Concepts Covered - 1

Calculation Of Changes In S For Different Process

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.

$\mathrm{dS = \frac{dq_{rev}}{T}\ or\ \Delta S = \frac{q_{rev}}{T}}$

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

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

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

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

$\therefore \mathrm{\Delta S = \frac{-w}{T}=\frac{2.303\ nRT\ log(\frac{V_2}{V_1})}{T}}$

$\therefore \mathrm{\Delta S = 2.303\ nR\ log(\frac{V_2}{V_1})=2.303\ nR\ log(\frac{P_1}{P_2})}$

2. Adiabatic reversible process

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

Note: Reversible adiabatic process is also called as Isentropic process

3. Isobaric process:

$\mathrm{\Delta S = 2.303\ nC_P\ log(\frac{T_2}{T_1})=2.303\ nC_P\ log(\frac{V_2}{V_1})}$

4. Isochoric process:

$\mathrm{\Delta S = 2.303\ nC_V\ log(\frac{T_2}{T_1})=2.303\ nC_V\ log(\frac{P_2}{P_1})}$

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

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

$\mathrm{\Delta S=\int \frac{nC_v dT+PdV}{T}=\int_{T_1}^{T_2} \frac{(nC_v dT)}{T} + \int_{V_1}^{V_2} \frac{(nR dV)}{V}}$

$\mathrm{\Delta S=n C_v ln(\frac{T_2}{T_1})+ nR ln(\frac{V_2}{V_1})}$

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

$\mathrm{\Delta S=n C_p ln(\frac{T_2}{T_1})+ nR ln(\frac{P_1}{P_2})}$

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.

$\mathrm{\Delta S _{\text {system }}=-\frac{ q }{ T _{1}}}$

$\mathrm{\Delta S _{\text {surroundings }}=+\frac{ q }{ T _{2}}}$

$\mathrm{\Delta S _{\text {process }}=\Delta S _{\text {system }}+\Delta S _{\text {surroundings }} =-\frac{ q }{ T _{1}}+\frac{ q }{ T _{2}}= q \frac{\left[ T _{1}- T _{2}\right]}{ T _{1} T _{2}}}$

$\\\because\mathrm{ T_1 > T_2} \\\\\therefore \mathrm{T_1 -T_2 >0}$

$\therefore \mathrm{\Delta S_{process} >0}$

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

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

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

$\therefore \mathrm{\Delta S = m\times s\times log \frac{T_2}{T_1}}$

9. Entropy Change in Mixing of Ideal Gases:

Suppose mole of gas 'P' and 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.

$\\ \mathrm{X}_{1}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}} \\ \\ {\mathrm{X}_{2}=\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}}$

$\Delta \mathrm{S} / \mathrm{mol}=-2.303 \mathrm{R} \frac{\left[\mathrm{n}_{1} \log _{10} \mathrm{X}_{1}\right.}{\mathrm{n}_{1}+\mathrm{n}_{2}}+\frac{\left.\mathrm{n}_{2} \log _{10} \mathrm{X}_{2}\right]}{\mathrm{n}_{1}+\mathrm{n}_{2}}$