## 9.47

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Blake_Katsev_2E
Posts: 113
Joined: Wed Sep 21, 2016 2:57 pm

### 9.47

9.47 Initially a sample of ideal gas at 323 K occupies 1.67 L
at 4.95 atm. The gas is allowed to expand to 7.33 L by two
pathways: (a) isothermal, reversible expansion; (b) isothermal,
irreversible free expansion. Calculate "Stot, "S, and "Ssurr for
each pathway.

I understand when calculating irreversible expansion you use nRln(v2/v1). But why when calculating free expansion irreversible work would you use this same increase in entropy? shouldnt you not use the ln for an irreversible process?

Ara Yazaryan 1E
Posts: 44
Joined: Wed Sep 21, 2016 2:59 pm

### Re: 9.47

The change in entropy is still the same, and you still would use ln in this case. Yet no work is done in free expansion, so the delta s of the surroundings is zero and q is zero. Only the internal entropy of the system increases in the irreversible process.

Jessica Huang 1M
Posts: 26
Joined: Wed Sep 21, 2016 2:59 pm

### Re: 9.47

I don't think delta S is dependent on whether or not the process is reversible or irreversible. In both cases volume increased for the gas, which means disorder of the gas molecules increases (delta S increases).