Question

10/ Calculate the change in entropy of gases in the following cases:
a) A 3.0 mol sample of an ideal gas expands reversibly and isothermally at 350 K until its volume doubled.
b) The temperature of 1.0 mol of an ideal monatomic gas is raised reversibly from 200 K to 300 K, with its volume kept constant.

Answer 1

Answer on Question #57122, Physics / Molecular Physics | Thermodynamics

10/ Calculate the change in entropy of gases in the following cases:

a) A 3.0 mol sample of an ideal gas expands reversibly and isothermally at $350\,\mathrm{K}$ until its volume doubled.

b) The temperature of 1.0 mol of an ideal monatomic gas is raised reversibly from $200\,\mathrm{K}$ to $300\,\mathrm{K}$, with its volume kept constant.

Solution:

a) For the expansion (or compression) of an ideal gas from an initial volume $V_0$ to a final volume $V$ at any constant temperature, the change in entropy is given by:

$$\Delta S = n R \ln \left(\frac {V}{V _ {0}}\right)$$

Here $n$ is the number of moles of gas and $R$ is the ideal gas constant.

$$\Delta S = 3.0 \cdot 8.31 \cdot \ln (2) = 17.28\,\frac{\mathrm{J}}{\mathrm{K}}$$

b) For heating or cooling of any system (gas, liquid or solid) at constant volume from an initial temperature $T_0$ to a final temperature $T$, the entropy change is

$$\Delta S = n C_V \ln \left(\frac {T}{T _ {0}}\right)$$

where the constant-volume heat capacity $C_V$ is constant and there is no phase change.

In the case of a monatomic gas

$$C_V = \frac {3}{2} R$$

where $R$ is the ideal gas constant

Thus,

$$\Delta S = \frac {3}{2} n R \ln \left(\frac {T}{T _ {0}}\right) = \frac {3}{2} \cdot 1.0 \cdot 8.31 \cdot \ln \left(\frac {300}{200}\right) = 5.05\,\frac{\mathrm{J}}{\mathrm{K}}$$

Answer: a) $17.28\,\frac{\mathrm{J}}{\mathrm{K}}$; b) $5.05\,\frac{\mathrm{J}}{\mathrm{K}}$

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