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Van der Waal's Equation - Derivation

The vander wa...

Question

The vander waal's equation of corresponding states for 1 mole of gas is [where $P_{c} =$ critical pressure, $T_{c} =$ critical temperature, $V_{c} =$ critical volume]
$[π = \frac{P}{P_{c}}, ϕ = \frac{V}{V_{c}} a n d θ = \frac{T}{T_{c}}]$

A

[π - \frac{3}{ϕ}] (3 ϕ - 1) = 8 R θ

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B

[π + \frac{3}{ϕ}] (3 ϕ - 1) = 8 R θ

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C

[π + \frac{3}{ϕ}] (3 ϕ + 1) = 8 R θ

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D

[π - \frac{3}{ϕ}] (3 ϕ - 1) = 8 R θ

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Solution

The correct option is A $[π - \frac{3}{ϕ}] (3 ϕ - 1) = 8 R θ$
$π P_{c} = P ϕ V_{c} = V θ T_{c} = T$
We know that,
$P_{c} = \frac{a}{27 b^{2}}$
$V_{c} = 3 b$
$T_{c} = \frac{8 a}{27 R b}$
Substituting the values, we get
$π \frac{a}{27 b^{2}} = P ϕ 3 b = V θ . \frac{8 a}{27 R b} = T$
The van der waal's equation for one mole of a gas:
$(P + \frac{a}{V^{2}}) (V - b) = R T$
Substituting values, we get
$[π - \frac{3}{ϕ}] (3 ϕ - 1) = 8 θ R$

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Similar questions

cot θ = \frac{1}{2}

sec ϕ = - \frac{5}{3}

θ \in (π, \frac{3 π}{2})

ϕ \in (\frac{π}{2}, π)

, then the value of

cot (θ - ϕ)

Q. Given that the reduced temperature,

θ = \frac{T}{T_{C}}

,the reduced pressure

π = \frac{F}{F_{C}}

, the reduced volume

ϕ = \frac{V}{V_{C}}

Thus, it can be said that the reduced equation of state may be given as:

Q. Given that the reduced temperature,

θ = \frac{T}{T_{C}}

the reduced pressure,

π = \frac{F}{F_{C}}

the reduced volume,

ϕ = \frac{V}{V_{C}}

. Thus, it can be said that the reduced equation of state may be given as :

\frac{P}{P_{c}} = P_{r}, \frac{T}{T_{c}} = T_{r}, a n d \frac{V_{m}}{V_{m, c}} = V_{r}

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}

V_{r}

3 {sin}^{2} θ + 2 {sin}^{2} ϕ = 1

3 sin 2 θ = 2 sin 2 ϕ

0 < θ < \frac{π}{2}

0 < ϕ < \frac{π}{2}

, then the value of

θ + 2 ϕ

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