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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 ?

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  • $\begingroup$ Thanks for reading the question. I found my mistake now. In the third step of "my attempt" $\left(\frac{\partial P}{ \partial V} \right)_T=0$ as pressure remains constant in that step. $\endgroup$
    – himat
    Commented Jan 3, 2017 at 18:22

1 Answer 1

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I am supposing you are working on the $\bf{VT}$ diagram, for otherwise it will be too trivial. So I suppose your first path is a straight line on the $VT$ diagram from $(V_0,T_0)$ to $(2V_0,4T_0)$, and your alternative pathway is a straight path on the diagram from $(V_0,T_0)$ to $(2V_0, 2T_0)$, followed by a horizontal path from $(2V_0,2T_0)$ to $(2V_0,4T_0)$. Notice that in the first process, both $V$ and $T$ change, while in the second process only $T$ changes. So you should have $$\Delta P =\int_{(V_0,T_0)}^{(2V_0,2T_0)}\left[\left(\frac{\partial P}{\partial V}\right)_TdV+\left(\frac{\partial P}{\partial T}\right)_VdT\right]+\int_{2T_0}^{4T_0}\left(\frac{\partial P}{\partial T}\right)_VdT$$ $$=\int_{(V_0,T_0)}^{(2V_0,2T_0)}\left[-\frac{P_0}{V}dV+\frac{P_0}{T}dT\right]+\int_{2T_0}^{4T_0}\frac{nR}{2V_0}dT$$ $$=-P_0\ln2+P_0\ln2+\frac{nR}{2V_0}2T_0$$ $$=\frac{nRT_0}{V_0}$$ $$=P_0$$

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