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Ideal Gas Equation

If P/Pc=Pr,...

Question

If $\frac{P}{P_{c}} = P_{r}, \frac{T}{T_{c}} = T_{r}, a n d \frac{V_{m}}{V_{m, c}} = V_{r}$
where
$P_{r}$ is reduced pressure, $P_{c}$ is critical pressure
$T_{r}$ is reduced temperature, $T_{c}$ is critical temperature
$V_{r}$ is reduced volume, $V_{c}$ is critical volume
then the equation of state (or van der Waals equation), only in terms of $P_{r}, T_{r}$ and $V_{r}$ is

(P_{r} + \frac{3}{V_{r}^{2}}) = 8 T_{r}

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(P_{r} + \frac{3}{V_{r}^{2}}) (3 V_{r} - 1)

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(P_{r} + \frac{3}{V_{r}^{2}}) (3 V_{r} - 1) = 4 T_{r}

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(P_{r} + \frac{3}{V_{r}^{2}}) (3 V_{r} - 1) = 8 T_{r}

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Solution

The correct option is D $(P_{r} + \frac{3}{V_{r}^{2}}) (3 V_{r} - 1) = 8 T_{r}$
The van der waals equation is
$[P + \frac{a}{V_{m}^{2}}] (V_{m} - b) = R T$
Substitute $P = P_{c} P_{r}, V = V_{c} V_{r} a n d T = T_{c} T_{r}$ in the above equation.
$[P_{c} P_{r} + \frac{a}{(V_{c} V_{r})^{2}}] (V_{c} V_{r} - b) = R T_{c} T_{r}$
Substitute $P_{c} = \frac{a}{27 b^{2}}, V_{c} = 3 b a n d T_{c} = \frac{8 a}{27 b R}$ in the above equation
$[\frac{a}{27 b^{2}} P_{r} + \frac{a}{(3 b V_{r})^{2}}] (3 b V_{r} - b) = R \frac{8 a}{27 b R} T r$
On rearranging the equation we get
$(P_{r} + \frac{3}{V_{r}^{2}}) (3 V_{r} - 1) = 8 T_{r})$

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If PPc=Pr,TTc=Tr,andVmVm,c=VrwherePr is reduced pressure, Pc is critical pressureTr is reduced temperature, Tc is critical temperatureVr is reduced volume, Vc is critical volumethen the equation of state (or van der Waals equation), only in terms of Pr,Tr and Vr is