## 6th Edition, 8.9

$\Delta U=q+w$

Douglas Nguyen 2J
Posts: 71
Joined: Fri Sep 28, 2018 12:15 am

### 6th Edition, 8.9

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?"

For this particular problem, when using converting P*deltaV from L*atm to J, why is it that we use the gas constant 8.206 x 10-2 L·atm·K-1·mol-1 instead of 1 L.atm = 101.325 J? Is it because the question states that it was an ideal gas being compressed?

Alexa Tabakian 1A
Posts: 38
Joined: Fri Sep 28, 2018 12:20 am

### Re: 6th Edition, 8.9

You still use the 101.325 J to convert from L x atm to Joules. You should get -150.97 Joules for work and then convert it to kJ which is -.15 kJ, this is work. You then add work and q which is 5.5 kJ. Your final answer should be 5.35 kJ.