Question

8.2. Prove that the slope on a TS diagram of (a) An isochoric curve is T/Cv. (b) An isobaric curve is T/C

Answer 1

(a.) For an ispchoric curve,

For isochoric (constant-volume,dV=0) processes, no work is done (work being PdV), so any energy change dU in the system must be in the form of heat transfer q.

Additionally, we can use the differential energy equation

$dU=TdS-PdV$

for constant volume,

$dU=TdS$

$TdS=T(\dfrac{\delta S}{\delta T})_VdT=C_VdT$

From above equation,

$\dfrac{dT}{dS}=\dfrac{T}{C_V}=slope$

(b.) we know from ideal gas equation-

$TdS=dH-VdP$ (i)

for constant pressure line

$VdP=0$

for an ideal gas

$dH=C_pdT$

From equation (i),

$TdS=C_pdT$

$\dfrac{dT}{dS}=\dfrac{T}{C_p}=slope$

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