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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}
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: $$\Delta S=nC_v\ln{(T_2/T_1)}+nR\ln{(V_2/V_1)}$$
