Question

the heat absorbed by a mole of an ideal gas in a quasistatic process in which temperature changes by dt and volume V changes by dv is given by
dQ=cdt+pdv
find the change in entropy of this gas in a quasistatic process which takes it from initial values of temperature and volume to the final values of temperature and volume . does your answer depend upon the process involved in going from initial to final state ?

Answer 1

Answer on Question #48261-Physics-Molecular Physics-Thermodynamics

The heat absorbed by a mole of an ideal gas in a quasistatic process in which temperature changes by $dT$ and volume $V$ changes by $dV$ is given by

$$\delta Q = C d T + \bar {P} d V,$$

where $C$ is its constant molar heat capacity at constant volume, and $\bar{P} = \frac{RT}{V}$ is the mean pressure.

Find the change in entropy of this gas in a quasistatic process which takes it from initial values of temperature $T_{i}$ and volume $V_{i}$ to the final values of temperature $T_{f}$ and volume $V_{f}$ . Does your answer depend upon the process involved in going from initial to final state?

Solution

$$d S = \frac {\delta Q}{T} = \frac {C d T + \bar {P} d V}{T} = \frac {C d T}{T} + \frac {\bar {P}}{T} d V.$$

But

$$\frac {\bar {P}}{T} = \frac {\frac {R T}{V}}{T} = \frac {R}{V}.$$

The change in entropy of this gas in a quasistatic process is

$$\Delta S = \int_ {T _ {i}} ^ {T _ {f}} \frac {C d T}{T} + \int_ {V _ {i}} ^ {V _ {f}} \frac {R d V}{V} = C \ln \frac {T _ {f}}{T _ {i}} + R \ln \frac {V _ {f}}{V _ {i}}.$$

As we can see it doesn't depend upon the process involved in going from initial to final state.

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