## 4B.5

$\Delta U=q+w$

JamieVu_2C
Posts: 108
Joined: Thu Jul 25, 2019 12:16 am

### 4B.5

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?

With the equation $\Delta$U = q + w, you find heat and work. For work, you get w= -P(external)Delta[/tex]V, which is w= -150 J. How is work negative?

Jonathan Gong 2H
Posts: 105
Joined: Sat Jul 20, 2019 12:16 am

### Re: 4B.5

Work is negative because the volume has increased at constant pressure... which means that the system is performing work on its surroundings and has lost energy as a result.

Jessica Chen 2C
Posts: 103
Joined: Thu Jul 11, 2019 12:17 am

### Re: 4B.5

I agree with what the previous reply said. In these types of problems, it's important to keep track of the signs of the values. A negative w value means the system is performing work on its surroundings, while a positive w value means the surroundings are performing work on the system. Likewise, a negative q value means heat is leaving the system, and a positive q value means heat is being added to the system.