### Entropy of a Irreversible Isothermal Expansion

##### Example:

The same piston-cylinder assembly is now loaded with a single
weight to yield the same initial pressure of 2 bar. The weight is
removed all at once. The initial and final P,V,T are the same as
the previous example. 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:

In order to calculate $\Delta S_{sys}$ we must always use the
**reversible** process, so even for the irreversible
process, we get

$\Delta S_{sys} = 5.76 J/K$

$\Delta S_{surr} = \int_{initial}^{final} \frac{\delta
Q_{surr}}{T}$

##### Recall:

Our first law analysis of this problem yields:

$Q_{sys}=-W = -(-\int P_E dV) = nP_2 \left (\hat V_2 - \hat
V_1 \right )$

$\Delta S_{surr} = -\int_{initial}^{final} \frac{nP_2 \left
(\hat V_2 - \hat V_1 \right )}{T}$

$\Delta S_{surr} = -\frac{nP_2}{T_1} \left (\hat V_2 - \hat
V_1 \right ) = -\frac{(1 mol)(1 bar)}{962 K}\left (0.08m^3/mol -
0.04m^3/mol \right ) = -4.16 J/K$

$\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}$

$\Delta S_{univ} = 5.75 J/K - 4.16 J/K = 1.59 J/K$

##### Outcome:

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