For an ideal gas, a constant pressure line and a constant volume line intersect at a point, in the Temperature (T) versus specific entropy (s) diagram. CP is the specific heat at constant pressure and CV is the specific heat at constant volume. The ratio of the slopes of the constant pressure and constant volume lines at the point of intersection is

The correct option/answer is:
$\frac{C_p-C_v}{C_v}$

Analysis of the Diagram:
Based on the provided T-s diagram (Temperature versus specific entropy), we observe two curves intersecting at a point:
1. A constant volume line labeled as v = C
2. A constant pressure line labeled as P = C
The diagram visually shows that the constant volume curve is steeper than the constant pressure curve at the point of intersection.

Step-by-Step Derivation of Slopes:
Using the fundamental thermodynamic relation (Gibbs equation) for a constant volume process (v = C):
$Tds=du+pdv$
Since $dv=0$, this simplifies to:
$Tds=du$
For an ideal gas, the change in internal energy is $du=C_{v}dT$. Substituting this gives:
$Tds=C_{v}dT$
Rearranging for the slope of the constant volume curve on the T-s diagram:
${(\frac{dT}{ds}}_v=\frac{T}{C_v}$

Similarly, using the second T-ds relation for a constant pressure process (P = C):
$Tds=dh-vdP$
Since $dP=0$, this simplifies to:
$Tds=dh$
For an ideal gas, the change in enthalpy is $dh=C_{p}dT$. Substituting this gives:
$Tds=C_{p}dT$
Rearranging for the slope of the constant pressure curve on the T-s diagram:
${(\frac{dT}{ds}}_p=\frac{T}{C_p}$

Determining the Ratio Relations:
The ratio of the constant volume slope to the constant pressure slope is:
$\frac{{\left(\frac{dT}{ds}\right)}_v}{{\left(\frac{dT}{ds}\right)}_p}=\frac{T/C_v}{T/C_p}=\frac{C_p}{C_v}$
The relative change in these slopes compared to the constant volume slope corresponds to:
$\frac{C_p-C_v}{C_v}$
This verifies that the correct matching option is indeed:
$\frac{C_p-C_v}{C_v}$