Question

A rigid tank, with a volume of 25 L, contains nitrogen at 355 K and 1.2 MPa. The tank is then cooled to 120 K. What work has been done and the heat transferred in this process? Data: Cv = 0.745 kJ/kgK, Cp = 1.042 kJ/kgK, MM(N2)=28 g/mol, R = 8.314 m^3 Pa/molK.

Use the ratio PV = nRT to determine the amount of nitrogen contained in the tank.

Answer #1

Initial volume = 25 L , temperature and pressure are 355 K and 1.2 Mpa

Final temperature = 120 K

Initial moles = PV/RT == 11.84 atm x 25 L / (0.082 x 355 )

= 10.17 moles

work done = 0

This is constant volume process.

C_{v} = 0.745 kJ/kg K = ( 0.745 kJ/kg K )*( 28 kg /kmol)
= 20.86 J/mol K

0.745 kJ/kgK

C_{p} - C_{v} = R

C_{p} = 20.86 + 8.314 = 29.174 J/mol K

Heat transferred = n C_{p}(T_{2}-T_{1}
)

= 10.17 moles x 29.174 J/mol K x (120 K - 355 K )

= - **69.7 kJ**

**Thus , heat transferred is - 69.7 kJ**

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