## 8.9

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

Jessica Wakefield 1H
Posts: 58
Joined: Thu Jul 27, 2017 3:00 am

### 8.9

An ideal gas in a cylinder was placed in a heater and gained 5.50 kJ of energy as heat. If the cylinder increased in volume from 345 mL to 1846 mL against an atmospheric pressure of 750 Torr during this process, what is the change in internal energy of the gas in the cylinder?

I was wondering why when converting to J after finding w in L.atm you have to multiply 1.48 L.atm by one R constant divided by another..

the solutions manual shows (1.48 L.atm)(8.314 J.K.mol^-1/0.08206 L.atm.K^-1.mol^-1) and I wasn't sure why you wouldn't use one R constant

Curtis Wong 2D
Posts: 62
Joined: Sat Jul 22, 2017 3:00 am

### Re: 8.9

Since we're trying to find the Internal Energy, our final answer should have energy units. By using these two R constants, our units would all cancel out to get joules as a final unit. That's why we use the two R constants.

904940852
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

### Re: 8.9

The answer is supposed to be in an energy unit, it's just a conversion to get the answer in Joules and cancel out any unnessasary units