Question

Prove that the slope of a constant-volume line is steeper than that for a constant-pressure line for a given state (point) on the T-s diagram.

Answer 1

Explanation:

We know that first T-ds equation

 Tds=$C_v$dT+Pdv

For constant volume process dv=0

⇒Tds=$C_v$dT

So $\dfrac{dT}{ds}=\dfrac{T}{C_v}$          ----(1)

We know that second T-ds equation

 Tds=$C_p$dT-vdP

For constant pressure process dP=0

⇒Tds=$C_p$dT

So $\dfrac{dT}{ds}=\dfrac{T}{C_p}$             ---(2)

We know that $C_p$ is always greater than $C_v$.

So from equation (1) and (2) we can say that slope of constant volume process will be always greater than constant pressure process on T-s diagram.