Chemical Engineering - Chemical Engineering Thermodynamics - Discussion
Discussion Forum : Chemical Engineering Thermodynamics - Section 3 (Q.No. 50)
50.
The change in __________ is equal to the reversible work for compression in steady state flow process under isothermal condition.
Discussion:
4 comments Page 1 of 1.
Raj Musale said:
6 years ago
According to the ideal gas equation.
Pv = RT.
By differentiating;
PdV + VdP = RdT.
For isothermal case;
PdV = -VdP.
Now dG = Vdp - SdT.
And for isothermal dG = Vdp = -PdV.
Pv = RT.
By differentiating;
PdV + VdP = RdT.
For isothermal case;
PdV = -VdP.
Now dG = Vdp - SdT.
And for isothermal dG = Vdp = -PdV.
(1)
Ajay Chauhan said:
8 years ago
Please Explain it.
Kapil said:
8 years ago
Shouldn't it be Helmholtz.
Mitul said:
8 years ago
G = -sdt + vdp , here condition is dt= 0.
Than G = vdp and it the reversible work done at steady state.
Than G = vdp and it the reversible work done at steady state.
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