## 8.31 heat released by Krypton gas

Kim Vu 2G
Posts: 29
Joined: Fri Sep 25, 2015 3:00 am

### 8.31 heat released by Krypton gas

Calculate the heat released by 5.025g of Kr at .4 atm as it cools from 97.6 degrees to 25 degrees at a)constant pressure, b)constant volume. Assume that krypton behaves as an ideal gas. For part a, why is the molar heat capacity 10.8J/mol*C and part b 12.5 J/mol*C? Does it have to do with ratio of specific heat?

JasmineAlberto4J
Posts: 73
Joined: Fri Sep 25, 2015 3:00 am

### Re: 8.31

to solve this problem you will have to use q=nC$\Delta$T where n is the number of moles and C is the heat capacity and $\Delta$ T is the change in temp.
However, gasses have special heat capacities that depend on whether the system is at a constant pressure or a constant volume. You can find these specific rules on pg 281
In this case we are dealing with a monatomic gas so we find the specific heat capacities:
at a constant pressure (Cp) by using 5/2R (R=Gas constant in J*mol^-1*K^-1) which gives you C=20.8J*^mol^-1*K^-1
at a constant volume (Cv) by using 3/2R (R=Gas constant in J*mol^-1*K^-1 ) which gives you C=12.5 J*^mol^-1*K^-1
*side note you will find that when you calculate your deltaT:
[Tf(+273)-Ti(+273] the +273's cancel out so you don't have to worry so much about K and C in these problems like you would in others.
(the +273 are to convert to K)
Last edited by JasmineAlberto4J on Fri Jan 15, 2016 11:37 am, edited 1 time in total.

Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

### Re: 8.31 heat released by Krypton gas

Where did you get 5/2R and what is R?

JasmineAlberto4J
Posts: 73
Joined: Fri Sep 25, 2015 3:00 am

### Re: 8.31 heat released by Krypton gas

Madeline Offerman 3G wrote:Where did you get 5/2R and what is R?

R is the gas constant and 5/2 R is the formula to find the heat capacity (C) of a monatomic gas