
How do you calculate the change in entropy for an isothermal process?
Answer
544.2k+ views
Hint:Entropy is defined as a variable used to calculate the randomness or disorder of the system. If the disorder in a system is high then it will show high entropy. At high temperature, the entropy also increases. The entropy order for solid, liquid and gas are- gas> liquid> solid.
Complete step-by-step answer:When the temperature is constant then the value of change in entropy will be-
1.\[\Delta S={{\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\left( \dfrac{\partial S}{\partial P} \right)}}_{T}}dP\]
2.\[\Delta S={{\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{\left( \dfrac{\partial S}{\partial V} \right)}}_{T}}dV\]
Now we use Maxwell relations that contain $T,P$ or $T,V$ as variables.
3.$dG=-SdT+VdP$
$G$ is the gibbs free energy
4.$dA=-SdT-PdV$
$A$ is Helmholtz free energy
Entropy as a function of temperature and pressure-
Firstly we will calculate the change in entropy for an isothermal process.
$\Rightarrow dG=-SdT+VdP$
Now from Maxwell relations do something similar with $V,T,P$ to get:
$\Rightarrow {{\left( \dfrac{\partial S}{\partial P} \right)}_{T}}=-{{\left( \dfrac{\partial V}{\partial T} \right)}_{P}}$
From ideal gas law, we calculate the value of volume
$\Rightarrow PV=nRT$
$\Rightarrow V=\dfrac{nRT}{P}$
Now, substitute the value of $V$
$\Rightarrow -{{\left( \dfrac{\partial V}{\partial T} \right)}_{P}}=-\dfrac{\partial }{\partial T}{{\left[ \dfrac{nRT}{P} \right]}_{P}}$
$\Rightarrow -\dfrac{1}{P}\dfrac{\partial }{\partial T}\left[ nRT \right]=-\dfrac{nR}{P}$
Now putting equation $\left( 1 \right)$
\[\Rightarrow \Delta S={{\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\left( \dfrac{\partial S}{\partial P} \right)}}_{T}}dP\]
$\Rightarrow -\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\left( \dfrac{\partial V}{\partial T} \right)}dP$
$\Rightarrow -\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\dfrac{nR}{P}}dP$
$\Rightarrow -nR\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\dfrac{1}{P}}dP$
$\Rightarrow -nR\ln \left( \dfrac{{{P}_{2}}}{{{P}_{1}}} \right)$
Additional information
Properties of entropy are:
Entropy is a state function that only depends on state not in path.
It is an extensive property.
It is a thermodynamic function.
Its SI unit is $J{{K}^{-1}}mo{{l}^{-1}}$ .
An isothermal process is defined as a process where the temperature of the system is constant. Examples of isothermal processes are – change of state through melting and evaporation, working of a refrigerator.
Note:In the above question, we have calculated the change in entropy for the isothermal process. Entropy increases with mass. When liquid or hard substances are dissolved in water we see an increase in entropy. But when gas is dissolved in water, entropy decreases.
Complete step-by-step answer:When the temperature is constant then the value of change in entropy will be-
1.\[\Delta S={{\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\left( \dfrac{\partial S}{\partial P} \right)}}_{T}}dP\]
2.\[\Delta S={{\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{\left( \dfrac{\partial S}{\partial V} \right)}}_{T}}dV\]
Now we use Maxwell relations that contain $T,P$ or $T,V$ as variables.
3.$dG=-SdT+VdP$
$G$ is the gibbs free energy
4.$dA=-SdT-PdV$
$A$ is Helmholtz free energy
Entropy as a function of temperature and pressure-
Firstly we will calculate the change in entropy for an isothermal process.
$\Rightarrow dG=-SdT+VdP$
Now from Maxwell relations do something similar with $V,T,P$ to get:
$\Rightarrow {{\left( \dfrac{\partial S}{\partial P} \right)}_{T}}=-{{\left( \dfrac{\partial V}{\partial T} \right)}_{P}}$
From ideal gas law, we calculate the value of volume
$\Rightarrow PV=nRT$
$\Rightarrow V=\dfrac{nRT}{P}$
Now, substitute the value of $V$
$\Rightarrow -{{\left( \dfrac{\partial V}{\partial T} \right)}_{P}}=-\dfrac{\partial }{\partial T}{{\left[ \dfrac{nRT}{P} \right]}_{P}}$
$\Rightarrow -\dfrac{1}{P}\dfrac{\partial }{\partial T}\left[ nRT \right]=-\dfrac{nR}{P}$
Now putting equation $\left( 1 \right)$
\[\Rightarrow \Delta S={{\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\left( \dfrac{\partial S}{\partial P} \right)}}_{T}}dP\]
$\Rightarrow -\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\left( \dfrac{\partial V}{\partial T} \right)}dP$
$\Rightarrow -\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\dfrac{nR}{P}}dP$
$\Rightarrow -nR\int\limits_{{{P}_{1}}}^{{{P}_{2}}}{\dfrac{1}{P}}dP$
$\Rightarrow -nR\ln \left( \dfrac{{{P}_{2}}}{{{P}_{1}}} \right)$
Additional information
Properties of entropy are:
Entropy is a state function that only depends on state not in path.
It is an extensive property.
It is a thermodynamic function.
Its SI unit is $J{{K}^{-1}}mo{{l}^{-1}}$ .
An isothermal process is defined as a process where the temperature of the system is constant. Examples of isothermal processes are – change of state through melting and evaporation, working of a refrigerator.
Note:In the above question, we have calculated the change in entropy for the isothermal process. Entropy increases with mass. When liquid or hard substances are dissolved in water we see an increase in entropy. But when gas is dissolved in water, entropy decreases.
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