# Heat added to Monatomic Gas at Constant Pressure

1. Nov 12, 2011

### KendrickLamar

1. The problem statement, all variables and given/known data

If 650 J of heat are added to 21 moles of a monatomic gas at constant pressure, how much does the temperature of the gas increase? (in Kelvins)

2. Relevant equations
U = nRT
Q=(5/2)nR(T2-T1)

3. The attempt at a solution

well how do you even know how much the temperature changes by if they dont give u an initial temperature? or is it just as simple as solving for delta T in which you just do

650J = (5/2)(21)( 8.31 J/mol.K )(Delta T)

is it really that simple or do i have to do something with the 21 moles and convert it or anything? and once i do get the Delta T must that be converted or something in order to have it end up in Kelvins? I ended up with 1.4899K im not sure if that's right though.

2. Nov 13, 2011

### ehild

It is correct. Do not write it out with more than 3 digits.

ehild

3. Nov 13, 2011

### Andrew Mason

Careful. For a monatomic ideal gas, internal energy, U = 3nRT/2
Note that T is in Kelvins, and Q is in Joules and R is in Joule/mol Kelvin. The molar heat capacity Cp (=5R/2) is temperature independent ie. it is the same for all T.
It is that simple.

AM

4. Nov 13, 2011

### KendrickLamar

wait so is it 3/2 or 5/2? cuz i have 2 equations from the formula sheet he gave us one says

Cp = 5/2 * R and one says Cv = 3/2 * R ? or in this case is it the 3/2*5/2 or something?

5. Nov 13, 2011

### ehild

Cp is the molar specific heat capacity at constant pressure, and it is 5/2 R for mono-atomic gases. You add heat at constant pressure, so Cp was correctly used.

ehild

6. Nov 13, 2011

### KendrickLamar

thank you guys, appreciate it alot

7. Nov 13, 2011

### Andrew Mason

Cp = 5R/2 and Cv=3R/2. But you had U = nRT. That is not correct. U = nCvT = (3/2)nRT.

AM