# SL: Entropy of a Rev/irreversible Isothermal Compression

### Entropy of a Rev/irreversible Isothermal Compression

##### Example:

The gas in the previous examples is compressed back isothermally and (ir)reversibly. Represent this on a PV and PT diagram. Find the total heat transfer, and the entropy change of the system, the surroundings, and the universe.

##### Note:

For both the reversible and irreversible case, $\Delta S_{sys}$ must use the reversible process.

$\Delta S_{sys_{(ir)rev}} = \int_{initial}^{final} \frac{\delta Q_{sys}}{T}$

##### Recall:

Our first law analysis of this problem (reversible) yields:

$Q=-W = -(-\int P_E dV) = nRT_1 \ln \left (\frac{P_2}{P_1} \right )$

$\Delta S_{sys_{(ir)rev}} = \int_{initial}^{final} \frac{nRT_1 \ln \left (\frac{P_2}{P_1} \right )}{T}$

$\Delta S_{sys_{(ir)rev}} = nR\ln \left (\frac{P_2}{P_1} \right ) = -(1 mol)(8.314 J/mol K) \ln (2) = -5.76 J/K$

$\Delta S_{surr_{rev}} = -\Delta S_{sys_{rev}} = 5.76 J/K$

$\Delta S_{univ_{rev}} = 0$

##### Recall:

In contrast, our first law analysis of this problem (irreversible) yields:

$Q_{sys}=-W = -(-\int P_E dV) = nP_1 \left (\hat V_1 - \hat V_2 \right )$

$\Delta S_{surr_{irrev}} = -\int_{initial}^{final} \frac{nP_1 \left (\hat V_1 - \hat V_2 \right )}{T}$

$\Delta S_{surr_{irrev}} = -\frac{nP_1}{T_1} \left (\hat V_1 - \hat V_2 \right ) = -\frac{(1 mol)(1 bar)}{962 K}\left (0.04m^3/mol - 0.08m^3/mol \right ) = 8.31 J/K$

$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr_{irrev}}$

$\Delta S_{univ} = -5.75 J/K + 8.31 J/K = 2.56 J/K$

##### Outcome:

Identify, formulate, and solve simple engineering problems (such as expansion/compression and power cycles)