Change of temperature of ideal gas

[hubber26]

Homework Statement

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)

Homework 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?

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

what is wrog please?

 

[Count Iblis]

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

 

[hubber26]

right. Thanks very much!