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Entropy Change - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

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  • Calculation Of Changes In S For Different Process is considered one the most difficult concept.

  • 6 Questions around this concept.

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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 :

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 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 \mathrm{\Delta S}.

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Calculation Of Changes In S For Different Process

Chemistry Part I Textbook for Class XI

Page No. : 183

Line : 40

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