Enthalpy change in ideal gas

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

Argon is heated from T1 to T2. Assume ideal gas behaviour. Calculate the energy input. Assume a constant volume, mass and constant specific heat capacity. Pressure variable

T1=523.15 K
T2=823.15 K
M= 0.03995 kg
m=50 kg
R=8.31
Cv=0.0125 kJ/(mol.K)

2. Relevant equations

Enthalpy change in gas
ΔH = ΔU + nRΔT

ΔU=CvΔT
n= m/M
ΔH=CvΔT+ (m/M)RΔT

3. The attempt at a solution

ΔH=((0.0125)*(300))+(50/0.03995)*( 8.31)*(300)…kj

Is this equal to the heat energy input?

Would this equal the amount of energy it would take to cool the gas, i.e. the reverse process?

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