The reduced temperature = θ = T T_ C The reduced pressure = π = P P_ C The reduced volume = ϕ = V V_ C Hence, it can be said that the reduced equation of state may be given as

Correct answer: A. ((π)/(3) + 1 ϕ ^ 2 )(3ϕ - 1) = (8)/(3)θ P = P _ c π = ( a 27b^ 2 )π V = V _ c ϕ = (3b)ϕ T = T _ c θ = ((8a)/(27Rb))θ Hence substituting in the van der Waal's equation (P + a V^ 2 )(V - b) = RT ⇒ (π + 3 ϕ ^ 2 )(3ϕ - 1) = 8θ This is reduced state equation (π + 3 ϕ ^ 2 )((3ϕ)/(8θ) - (1)/(8θ)) = 1 ((π)/(3) + 1 ϕ ^ 2 )(3ϕ - 1) = (8)/(3)θ

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: States of Matter

NTA Abhyas 2020 · Chemistry · States of Matter

The reduced temperature = θ = T T_ C The reduced pressure = π = P...

The reduced temperature

= \text{\theta} = \dfrac{\text{T}}{{\text{T}_{\text{C}}}

The reduced pressure

= \text{\pi} = \dfrac{\text{P}}{{\text{P}_{\text{C}}}

The reduced volume

= \text{ϕ} = \dfrac{\text{V}}{{\text{V}_{\text{C}}}

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

Options

A $\left(\dfrac{\pi}{3} + \dfrac{1}{{ϕ}^{2}}\right)\left(3ϕ - 1\right) = \dfrac{8}{3}\theta$
B $\left(\dfrac{\pi}{3} + \dfrac{1}{ϕ}\right)\left(ϕ - 1\right) = \dfrac{8}{3}\theta$
C $\left(\dfrac{\pi}{4} + \dfrac{1}{ϕ}\right)\left(3\theta - 1\right) = \dfrac{8}{3}ϕ$
D $\left(\dfrac{\pi}{3} + \dfrac{1}{ϕ}\right)\left(3ϕ - 1\right) = \dfrac{8}{3}\theta$

Correct answer

A. $\left(\dfrac{\pi}{3} + \dfrac{1}{{ϕ}^{2}}\right)\left(3ϕ - 1\right) = \dfrac{8}{3}\theta$

Step-by-step solution

P = {P}_{c}\pi = \left(\dfrac{a}{27b^{2}}\right)\pi

V = {V}_{c}ϕ = \left(3b\right)ϕ

T = {T}_{c}\theta = \left(\dfrac{8a}{27Rb}\right)\theta

Hence substituting in the van der Waal's equation

\left(P + \dfrac{a}{V^{2}}\right)\left(V - b\right) = RT

<br/><br/>

⇒ \left(\pi + \dfrac{3}{{ϕ}^{2}}\right)\left(3ϕ - 1\right) = 8\theta

<br/><br/>This is reduced state equation<br/><br/> \left(\pi + \dfrac{3}{{ϕ}^{2}}\right)\left(\dfrac{3ϕ}{8\theta} - \dfrac{1}{8\theta}\right) = 1

\left(\dfrac{\pi}{3} + \dfrac{1}{{ϕ}^{2}}\right)\left(3ϕ - 1\right) = \dfrac{8}{3}\theta

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