An fixed amount of ideal gas in a container is subjected to the following procedure:
$(P_0,V_0,T_0) \rightarrow (2P_0,2V_0,4T_0)$
$P_0$ is initial pressure. $V_0$ is initial volume. $T_0$ is initial temperature.
I need to prove that if do the same process through an alternate pathway the pressure change will be same as the original pathway, as $P$ is a state function.
The alternate pathway is $$(P_0,V_0,T_0) \rightarrow (P_0,2V_0,2T_0) \rightarrow (2P_0,2V_0,4T_0)$$
My attempt:
$$dP=\left(\frac{\partial P}{ \partial V} \right)_T dV + \left(\frac{\partial P}{ \partial T}\right)_V dT $$
$$\implies \Delta P = \int_{V_0}^{2V_0} -\frac{nRT_0}{V^2} dV + \int_{2T_0}^{4T_0} \frac{nR}{2V_0}dT$$
$$\implies \Delta P = \left[\frac{nRT_0}{V}\right]_{V_0}^{2V_0} + \left[\frac{nRT}{2V_0}\right]_{2T_0}^{4T_0}$$
$$ \implies \Delta P = -\frac{nRT_0}{2V_0}+\frac{nRT_0}{V_0} = \frac{nRT_0}{2V_0} = \frac{P_0}{2}$$
However, in the alternative pathway too, the net change in pressure should have come out to be $P_0$ and not $\dfrac{P_0}{2}$. Where am I going wrong ?