On a P-V diagram of an ideal gas, suppose a reversible adiabatic line intersects a reversible isothermal line at point A. Then at a point A, the slope of the reversible adiabatic line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}}$$  and the slope of the reversible isothermal line $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}}$$  are related as (where, $${\text{y}} = \frac{{{{\text{C}}_{\text{p}}}}}{{{{\text{C}}_{\text{v}}}}}$$  )

A. $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}} = {\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}}$$

B. $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}} = {\left[ {{{\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)}_{\text{T}}}} \right]^{\text{y}}}$$

C. $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}} = {\text{y}}{\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}}$$

D. $${\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{S}}} = \frac{1}{{\text{y}}}{\left( {\frac{{\partial {\text{P}}}}{{\partial {\text{V}}}}} \right)_{\text{T}}}$$

Answer: Option C

Solution(By Examveda Team)

For an adiabatic process → $$P{V^y}$$ = constant
For an isothermal process → $$PV$$ = constant
So, slope for adiabatic process in $$PV$$ plane is $$\frac{{dp}}{{dv}} = - y\frac{p}{v}$$
Slope for isothermal process is $$\frac{{dp}}{{dv}} = - \frac{p}{v}$$
Hence, $${\left( {\frac{{\partial p}}{{\partial v}}} \right)_S} = - y{\left( {\frac{{\partial p}}{{\partial v}}} \right)_T}$$