# Change of temperature of ideal gas

1. May 7, 2009

### hubber26

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

We have 1m3 of Ar gas M=39.948g/mol, pressure=101300Pa, temperature T=298K. We put the energy Q=1000J in that gas. Whats the temperature after we put 103J. (c=0.5203x103)

2. Relevant equations

Im not interested in solution. I tried to solve it on two ways using formula for internal energy of ideal gas $$\Delta$$U = $$\frac{3}{2}$$nR$$\Delta$$T and the formula Q = mc$$\Delta$$T

However solutions are not the same, why?

3. The attempt at a solution

pV=nRT => n = pV / RT = 40.89mol
m = Mn = 1633g = 1.633kg

$$\Delta$$U = $$\frac{3}{2}$$nR$$\Delta$$T
thus $$\Delta$$T = Q / $$\frac{3}{2}$$nR = 1000 / (1.5*40.89*8.314) = 1.961°K

however
Q = mc$$\Delta$$T
thus $$\Delta$$T = Q / mc = 1000 / (1.633*0.5203*103) = 1.177°K

2. May 7, 2009

### Count Iblis

You took c to be c_p while you need c_v here.

3. May 7, 2009

### hubber26

right. Thanks very much!