$5.9.5.$ Calculate the entropy increment of hydrogen of mass $m$ as it passes from volume $V_{1}$ and temperature $T_{1}$ to volume $V_{2}$ and temperature $T_{2}$ if the gas: a) heats up at a constant volume $V_{1}$ and then expands isothermically; b) expands at a constant temperature $T_{1}$ to volume $V_{2}$, then heats up at a constant volume; c) adiabatically expands to $V_{2}$ volume and then heats up at a constant volume.
As entropy is a function of state, then it is independent of path and the answer to all questions will be the same.
$$\Delta S = \int_{1}^{2}\frac{\delta Q}{T}$$
From the First law of thermodynamics:
$$\delta Q= dU + \delta A$$
$$\delta Q= \nu C_{V}dT + pdV$$
Dividing both sides at $dT$ will give us the entropy:
$$dS= \nu C_{V}\frac{dT}{T} + p\frac{dV}{T}$$
From the ideal gas law:
$$pV = \nu RT$$
$$\frac{p}{T} = \frac{\nu R}{V}$$
This yields:
$$dS= \nu C_{V}\frac{dT}{T} + \nu R\frac{dV}{V}$$
Integrating and substituting $C_{V} = \frac{i}{2}R$:
$$\Delta S = \frac{i}{2}\nu R\int_{1}^{2}\frac{dT}{T} + \nu R\int_{1}^{2}\frac{dV}{V}$$
$$\Delta S = \frac{3}{2}\nu R\cdot ln\frac{T_{2}}{T_{1}} + \nu R\cdot ln\frac{V_{2}}{V_{1}}$$
$$\Delta S = \nu R\left ( ln\left (\frac{T_{2}}{T_{1}} \right )^{\frac{3}{2}} + ln\frac{V_{2}}{V_{1}}\right)$$
$$\Delta S = \frac{m}{M}R\left ( ln\left (\frac{T_{2}}{T_{1}} \right )^{\frac{3}{2}}\frac{V_{2}}{V_{1}}\right)$$
$$\boxed{\Delta S = \frac{m}{M}R\left ( ln\left (\frac{T_{2}}{T_{1}} \right )^{\frac{3}{2}}\frac{V_{2}}{V_{1}}\right)}$$