- #1

- 12

- 0

## Homework Statement

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)

## Homework Equations

Enthalpy change in gas

ΔH = ΔU + nRΔT

ΔU=CvΔT

n= m/M

ΔH=CvΔT+ (m/M)RΔT

## 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?