## 8.11-is part A isothermal too? [ENDORSED]

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

sofiakavanaugh
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
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### 8.11-is part A isothermal too?

Hi,

I was reviewing my thermo homework and was taking a look at 8.11 where it asks

"A piston confines 0.200 mol Ne(g) in 1.20 L at 25 'C.
Two experiments are performed. (a) The gas is allowed to
expand through an additional 1.20 L against a constant
pressure of 1.00 atm. (b) The gas is allowed to expand
reversibly and isothermally to the same final volume. Which
process does more work?"

I understand how both are calculated, and in part a we use w=-pdeltaV. My question is, we didn't use temperature in the first part to calculate work, but it doesn't specify that it is isothermal either. Are situations like these isothermal, why or why not? And if the are then why did they specify that it is isothermal in part b for the reversible expansion but not for part a? Is there a reason behind this?

Thanks!

Danah Albaaj 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: 8.11-is part A isothermal too?  [ENDORSED]

The irreversible equation is not isothermal because the -nRT within the equation for reversible, isothermal expansion is in a way replacing the Pex from the irreversible equation. The isothermal is used to make -nRT a constant during the integration of the volume of the reversible expansion (pg 265 of the textbook shows the step-by-step of this with a more in-depth explanation if you would like to see it). The main focus of these calculations is how the difference in changing the pressure or keeping the pressure constant pressure will affect the results.