Can anyone help me with the derivation for the formula of entropy?

I have this formula in the chapter of thermodynamics:


$$\Delta S = n C\_v \ln\left(\frac{T\_2}{T\_1}\right) + n R \ln\left(\frac{V\_2}{V\_1}\right),$$
where $\Delta S$ is entropy, $n$ moles, $C\_v$ molar specific heat capacity at constant volume, and $R$ universal gas constant.


I'm curious how it got derived. Can anybody please derive this formula for me?




For an ideal gas, you can derive it this way:


\begin{align}
\Delta S &= \int\_{T\_1,V\_1}^{T\_2,V\_2}{\frac{\delta Q\_\mathrm{rev}}{T}} \\
&= \int\_{T\_1}^{T\_2}{\left(\frac{\delta Q\_\mathrm{rev}}{T}\right)\_V} + \int\_{V\_1}^{V\_2}{\left( \frac{\delta Q\_\mathrm{rev}}{T} \right)\_T }\\
&= \int\_{T\_1}^{T\_2}{\left(nC\_V \frac{\mathrm{d}T}{T}\right)\_V} +
\int\_{V\_1}^{V\_2}{\left(\frac{p \cdot \mathrm{d}V}{T}\right)\_T} &( \delta Q\_\mathrm{rev})\_T = (p.\mathrm{d}V)\_T\\
&= nC\_V \int\_{T\_1}^{T\_2}{\left( \frac{\mathrm{d}T}{T}\right)\_V} +
nR \cdot \int\_{V\_1}^{V\_2}{\left(\frac{ \mathrm{d}V}{V}\right)\_T} &(p=\frac{nRT}{V})\\
&= nC\_V \ln {\left(\frac{T\_2}{T\_1}\right)} +
nR \ln {\left(\frac{V\_2}{V\_1}\right)} \\
\end{align}


If we consider an isobaric reversible thermal expansion instead, it would be:


\begin{align}
\Delta S &= nC\_V \ln {\left(\frac{T\_2}{T\_1}\right)} +
nR \ln {\left(\frac{V\_2}{V\_1}\right)} \\
&= nC\_V \ln {\left(\frac{T\_2}{T\_1}\right)} +
nR \ln {\left(\frac{T\_2}{T\_1}\right)}\\
&= n(C\_V + R) \ln {\left(\frac{T\_2}{T\_1}\right)} \\
&= nC\_p \ln {\left(\frac{T\_2}{T\_1}\right)}
\end{align}


 5

For an ideal gas undergoing a reversible process, the 1st law of thermodynamics tells us that $$cU=nC\_vdT=dq-PdV=dq-\frac{nRT}{V}dV$$Dividng both sides of this equation by T then gives: $$\frac{dq}{T}=nC\_v\frac{dT}{T}+nR\frac{dV}{V}$$Each term in this equation is an exact differential, with the left side representing dS. Therefore integrating, we obtain: $$\Delta S=nC\_v\ln{(T\_2/T\_1)}+nR\ln{(V\_2/V\_1)}$$


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