Two processes isobaric and isochoric are represented on T-s diagram. They are starting from the same point. Out of these process which shall have a higher slope?

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UPRVUNL JE ME 2016 Official Paper Shift 2

Option 3 : isochoric

__Concept:__

Combined equations of the first and second law of thermodynamics

Tds = du + Pdv

Tds = dh – vdP

These equations are applicable for both reversible and irreversible process and for the closed and open system as well.

du = CvdT

dh = CpdT

Cv = specific heat at constant volume, Cp = specific heat at constant pressure

From second equation

Tds = dh – vdP

For constant pressure, dP = 0 & dh = CpdT

Tds = CpdT

\({\left. {\frac{{dT}}{{dS}}} \right|_{p = c}} = \frac{T}{{{C_p}}}\)

Hence on the T-S diagram, the slope of the constant pressure line = T/Cp

From first equation

Tds = du + Pdv

For constant volume, dv = 0

And, du = CvdT

∴ 1st equation becomes: Tds = CvdT

\({\left. {\frac{{dT}}{{dS}}} \right|_{v = c}} = \frac{T}{{{C_v}}}\)

Hence on the T-S diagram, the slope of the constant pressure line = T/C_{v}

As, C_{p }> C_{v} , T/Cv > T/Cp , Hence the slope of the Isochoric curve will be more than the slope of the Isobaric Curve on TS plane.