## Ch.8 Exercise 31

Ariana_Kermani_2G
Posts: 10
Joined: Fri Jul 15, 2016 3:00 am

### Ch.8 Exercise 31

Calculate the heat released by 5.025g of Kr(g) at 0.400 atm as it cools from 97.6 degrees Celsius to 25.0 degrees Celsius at (a) constant pressure and (b) constant volume. Assume that krypton behaves as an ideal gas.

What are the equations used to solve this problem and why do we need to use two different molar heat capacities?

Preston_Dang_1B
Posts: 22
Joined: Wed Sep 21, 2016 2:59 pm

### Re: Ch.8 Exercise 31

You will need to use the equation and plug in the values you are given, the mass, heat capacity constant for Krypton, and the change in temperature. The reason for there being two different heat capacities is because the heat capacity constant for any substance is different depending on the different conditions where this process is taking place. In this case, the conditions are constant pressure and constant volume and have different values, respectively.

Athena Dong 2A
Posts: 11
Joined: Sat Jul 09, 2016 3:00 am

### Re: Ch.8 Exercise 31

Also, because Kr(g) is a monatomic gas, I believe we can use the molar heat capacities at constant volume ( 3/2 R ) and constant pressure ( 5/2 R ) for the 'C' part of the equation. R here refers to the ideal gas constant; the 8.314 J.K^-1.mol^-1 value works well because it corresponds to the problem's given information.

There's an explanation regarding how we reach those molar heat capacity values in section 8.10 of the book, but I can also explain here if you want!