A quantity dz = Mdx + Ndy would be an exact differential, and hence a thermodynamic property, if

\left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y

(a) Here M=\frac{v}{T}=\frac{R}{p}

    (pv = RT)

\begin{aligned}&M=\frac{v}{T}=\frac{R}{p} \\&N=\frac{p}{T}=\frac{R}{v}\end{aligned}

x = p and y = v

\begin{aligned}\left(\frac{\partial M}{\partial y}\right)_x &=\left[\left(\frac{\partial}{\partial v}\right)\left(\frac{R}{p}\right)\right]_p=0 \\\left(\frac{\partial N}{\partial x}\right)_y &=\left[\left(\frac{\partial}{\partial p}\right)\left(\frac{R}{v}\right)\right]_v=0\end{aligned}

Since \left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y,the given expression is an exact differential, and hence thermodynamic property.

(b) Here  M=\frac{1}{T}

N=\frac{p}{T}

x = T and y = v

\begin{aligned}\left(\frac{\partial M}{\partial y}\right)_x &=\left[\left(\frac{\partial}{\partial v}\right)\left(\frac{1}{T}\right)\right]_T=0 \\\left(\frac{\partial N}{\partial x}\right)_y &=\left[\left(\frac{\partial}{\partial T}\right)\left(\frac{R}{v}\right)\right]_v=0\end{aligned}

\left(\frac{p}{T}=\frac{R}{v}\right)

Since \left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y, the given expression is an exact differential, and hence a thermodynamic property.

(c) Here M=\frac{1}{T}

N=\frac{p}{v}

x = T and y = v

\begin{aligned}&\left(\frac{\partial M}{\partial y}\right)_x=\left[\left(\frac{\partial}{\partial v}\right)\left(\frac{1}{T}\right)\right]_T=0 \\&\left(\frac{\partial N}{\partial x}\right)_y=\left[\left(\frac{\partial}{\partial T}\right)\left(\frac{p}{v}\right)\right]_v \neq 0\end{aligned}

Since \left(\frac{\partial M}{\partial y}\right)_x \neq\left(\frac{\partial N}{\partial x}\right)_y, the given expression is not an exact differential, and hence not a thermodynamic property

(d) Here M=\frac{1}{T}

N=\frac{v}{T}

x = T and y = p

\begin{aligned}\left(\frac{\partial M}{\partial y}\right)_x &=\left[\left(\frac{\partial}{\partial p}\right)\left(\frac{1}{T}\right)\right]_T=0 \\\left(\frac{\partial N}{\partial x}\right)_y &=\left[\left(\frac{\partial}{\partial T}\right)\left(\frac{R}{P}\right)\right] p=0\end{aligned}

\left(\because \frac{v}{T}=\frac{R}{p}\right)

Since \left(\frac{\partial M}{\partial y}\right)_x=\left(\frac{\partial N}{\partial x}\right)_y, the given expression is an exact differential, and hence a thermodynamic property.