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enter image description here

The given Z vs P curve for 1 mole of a gas at 400K starts at Z=1 & P=0.

The slope at the point when the curve again intersect Z=1 is 0.005. The critical temperature of the gas is 500K.

My approach:

From the van der Waals' equation,

$$(P+\frac{a}{V^2})(V-b)=RT$$

On expanding, $$\frac{PV}{RT}=Z=1-\frac{a}{VRT}+\frac{Pb}{RT}+\frac{ab}{V^2RT}$$

And $$ PV=ZRT $$ $-(1)$

By both equations, $$Z=1-\frac{a}{VRT}+\frac{Zb}{V}+\frac{ab}{V^2RT}$$ $-(2)$

Differentiating with respect to Z for Equation (1)

$$ PV'+VP'=RT$$

Equation (2)

$$ 1= \frac{aV'}{V^2RT}+\frac{b}{V}-\frac{bV'}{V^2}-\frac{2abV'}{V^3RT}$$

& $$T_c=\frac{8a}{27Rb}$$

Now I am stuck, I don't even know if I am moving in the right direction because things are becoming too complex to solve.

Where,

a, b = van der Waals' gas constants

Z = Compressibility Factor

V = Volume of gas when curve again intersects Z = 1

R = Universal gas constant

T = 400K

$T_c$ = Critical Temprature

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  • $\begingroup$ @Poutnik yes angle a pressure dependent and it is specific angle on the point when curve intersects Z=1. $\endgroup$
    – Leibniz-Z
    Commented Jan 3, 2023 at 18:56
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    $\begingroup$ I mean something else. For given pressure, angle depends on pressure unit. Tangens of angle is 1000x bigger for kPa than for Pa. $\endgroup$
    – Poutnik
    Commented Jan 3, 2023 at 19:00
  • $\begingroup$ Yes, you are right. But I couldn't edit that one because it is an image from web. $\endgroup$
    – Leibniz-Z
    Commented Jan 3, 2023 at 19:02
  • $\begingroup$ You need Z=f(p), not Z=(p,V(p)). But as you are interested in region near Z=1, you can do substitution V=RT/p. $\endgroup$
    – Poutnik
    Commented Jan 3, 2023 at 19:17
  • $\begingroup$ Can you please explain briefly in the answer? $\endgroup$
    – Leibniz-Z
    Commented Jan 3, 2023 at 19:34

2 Answers 2

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You were going in the right direction, But I has an extra step.

You already had proved the following equations.

$$ PV'+VP'=RT-(1)$$$$ 1= \frac{aV'}{V^2RT}+\frac{b}{V}-\frac{bV'}{V^2}-\frac{2abV'}{V^3RT}-(2)$$$$T_c=\frac{8a}{27Rb}-(3)$$$$\frac{PV}{RT}=Z=1-\frac{a}{VRT}+\frac{Pb}{RT}+\frac{ab}{V^2RT}-(4)$$

Substituting values $T_c=500K$, $T=400K$, $R=0.08$ (it will make calculation easier), $P'= \frac{dP}{dZ} = \frac{1}{0.05}=200 $ (at POI) in equation $(1),(2),(3)$ gives,

$$ \frac{32V'}{V}+200V=32-(A)$$$$ 1= \frac{aV'}{32V^2}+\frac{b}{V}-\frac{bV'}{V^2}-\frac{abV'}{16V^3}-(B)$$$$\frac{a}{b}=135-(C)$$ Now, if you look at the equation $(4)$. If you you put $Z=1$

$$\frac{a}{VRT}=\frac{Pb}{RT}+\frac{ab}{V^2RT}$$

Canceling out $ \frac{1}{VRT} $ and Substituting $P=RT/V$ gives,

$$a=32b+\frac{ab}{V}⟹\frac{a}{b}=32+\frac{a}{V}⟹135=32+\frac{a}{V}$$

So,

$$\frac{a}{V}=103-(D)$$

So, we're done.

By equation $(A),(B),(C),(D)$,

$$a= \frac{46532}{2575}≈18.07$$ $$b=\frac{46532}{347625}≈0.134$$ $$V=\frac{46532}{265225}≈0.175$$ $$V'=-\frac{47648768}{2813772025}≈-0.017$$

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First, there is a mistake in the fifth expression. It should be written $$PV' + P'V = RT$$ Second, it is not clear what Siddharth is looking for. If he wants a way of calculating $a$ with other van der Wals parameters, the simplest way is to draw $a$ from the final formula giving the critical temperature : $$\ce{a = \frac{27}{8}bRT_c}$$ Here the curve $Z(P)$ is not needed.

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  • $\begingroup$ Actually b is unknown. But this gives the relation $\frac{a}{b}≈135$. That I've used. $\endgroup$
    – Leibniz-Z
    Commented Jan 5, 2023 at 17:46
  • $\begingroup$ Can you tell what is the mistake in fifth expression, So I could correct it. $\endgroup$
    – Leibniz-Z
    Commented Jan 5, 2023 at 17:48
  • $\begingroup$ @Siddharth. You have written $PV' + V'P = ...$ If this had been correct, it could have been rewritten $PV' + V'P = 2 PV'$. But, care ! The position of the second "prime" is wrong. You should have written $PV' + P'V = ..$ $\endgroup$
    – Maurice
    Commented Jan 5, 2023 at 18:10
  • $\begingroup$ Oh yes! Sorry that was typing mistake. How could I further proceed to the question? $\endgroup$
    – Leibniz-Z
    Commented Jan 5, 2023 at 18:34
  • $\begingroup$ What are you looking for ? What is the question ? $\endgroup$
    – Maurice
    Commented Jan 5, 2023 at 19:58

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