Thermodynamic processes

$$ds=C_v\frac{dT}{T}+R\frac{d\nu }{\nu }$$

$$d\nu =\frac{\nu }{R}ds-C_v\frac{\nu }{RT}dT=\frac{\nu }{R}ds-\frac{C_v}{p}dT$$

for an isentropic process ($ds=0$), $d\nu <0$ for positive values of $dT$.

$$ds=C_p\frac{dT}{T}-R\frac{dp}{p}$$

$$dp=-\frac{p}{R}ds+C_p\frac{p}{RT}dT=-\frac{p}{R}ds+C_p\rho dT$$

for an isentropic process ($ds=0$), $dp>0$ for positive values of $dT$.

Since $\nu$ decreases with temperature and pressure increases with temperature for an isentropic process, we can see from Eqn.~\ref{eqn:process:dnu} that $d\nu$ will be greater at lower temperatures and thus isochores (lines of constant specific volume) will be closely spaced at low temperatures and more sparse at higher temperatures. For an isochore $dv=0$ which implies

$$0=\frac{\nu }{R}(ds-C_v\frac{dT}{T})\Rightarrow \frac{dT}{ds}=\frac{T}{C_v}$$

and thus we can see that the slope of an isochore in a $T-s$-diagram is positive and that the slope increases with temperature.

In analogy, we can see that an isobar ($dp=0$) leads to the following relation

$$0=\frac{p}{R}(C_p\frac{dT}{T}-ds)\Rightarrow \frac{dT}{ds}=\frac{T}{C_p}$$

and consequently isobars will also have a positive slope that increases with temperature in a $T-s$-diagram. Moreover, isobars are less steep than ischores as $C_p>C_v$.