# SL: Entropy of a Reversible Isothermal Expansion

### Entropy of a Reversible Isothermal Expansion

##### Example:

Consider a reversible, isothermal (T=962$K$), expansion from a pressure of 2 bar and a specific volume of 0.04$m^3/mol$ to a pressure of 1 bar and a specific volume of 0.08$m^3/mol$. Calculate the change in entropy of the system, the surroundings, and the universe.

$\Delta S_{sys} = \int_{initial}^{final} \frac{\delta Q}{T}$

##### Recall:

Out first law analysis of this problem yields:

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

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

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

##### Note:

For a reversible process the entropy change of the universe does not change. Thus, $\Delta S_{sys} = -\Delta S_{surround}$

##### Outcome:

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