Question

How can we calculate $\Delta S$ for an isothermal process?

Answer 1

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, entropy also increases.

Complete answer:
When temperature is constant then the value of change in entropy will be
$ \Delta S=\int^{P_{2}}_{P_{1}} (\dfrac{\delta S}{\delta P} )_{T}\ dP$
$ \Delta S=\int^{V_{2}}_{V_{1}} (\dfrac{\delta S}{\delta V} )_{T}\ dV$
Now we use Maxwell relation that contains T, P or T, V as variables.
$ dG=-SdT+VdP$
G is the Gibbs free energy.
$ dA=-SdT-PdV$
A is Helmholtz free energy.
Entropy is a function of temperature and pressure.
Now, let us calculate the change in entropy for an isothermal process.
$ dG=-SdT+VdP$
Now from Maxwell relations do something similar with V, T, P to get:
$ (\dfrac{\delta S}{\delta P} )_{T}=-(\dfrac{\delta V}{\delta T} )_{P}$
From Ideal gas law, we calculate the value of volume
$ PV=nRT$
$ V=\dfrac{nRT}{P}$
Now, substitute the value of V
$ -(\dfrac{\delta V}{\delta T} )_{P}=-\dfrac{\delta }{\delta T} [\dfrac{nRT}{P} ]_{P}$
$ -\dfrac{1}{P} \dfrac{\delta }{\delta T} [nRT]= -\dfrac{nR}{P}$
Now putting the first equation, we get
$ \Delta S=\int^{P_{2}}_{P_{1}} (\dfrac{\delta S}{\delta P} )_{T}\ dP$
$ -\int^{P_{2}}_{P_{1}} (\dfrac{\delta V}{\delta T} )dP$
$ -\int^{P_{2}}_{P_{1}} \dfrac{nR}{P} dP$
$ \Delta S =-nR\ \text{In} (\dfrac{P_{2}}{P_{1}} )$

Note:
In the above question, we have calculated the charge 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.