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A closed system containing an ideal gas undergoes an isentropic expansion process followed by a constant volume heat rejection process. Both processes are reversible. Assuming constant specific heat, write an equation for heat transfer and work for each process.

Respuesta :

Answer:

For isentropic process : Q= 0 ,[tex]\Delta W=-mC_v(T_2-T_1)[/tex]

For constant volume : [tex]\Delta Q=mC_v(T_3-T_2)[/tex],W=0

Explanation:

Given that:

System is closed

In first process ideal gas goes an isentropic expansion and then constant volume heat rejection.

We know that from first law of thermodynamic

Q = ΔU +  W

For process 1-2:

Process is adiabatic it means that Q=0

So  Q = ΔU +  W

0 =  ΔU +  W

W= -ΔU

We know that internal energy for ideal gas

[tex]\Delta U=mC_v(T_2-T_1)[/tex]

So work W

[tex]\Delta W=-mC_v(T_2-T_1)[/tex]

For process 2-3:

Process is constant volume so work transfer will be zero ,W=0

Q = ΔU +  W

Q = ΔU +  0

Q = ΔU

[tex]\Delta U=mC_v(T_3-T_2)[/tex]

[tex]\Delta Q=mC_v(T_3-T_2)[/tex]

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