One mole of triatomic vapours of an unknown substance effuses (4)/(3) time faster than 1 mole of O_2 under identical conditions. If the density of unknown vapours at pressure P and temperature T is d , which of the

Correct answer: D. z (compressibility factor) = 18 P dRT aligned & R_ vap R_ O_2 = 32 M_ vap =(4)/(3) & M_ vap =(32 × 9)/(16)=18 & VD=9 ; Density at NTP & =(18)/(22.4)=0.8035 gm / L & Z=(P V)/(n R T)=(P M)/(d R T)= 18 P d R T aligned

JEE Advanced

Exam: JEE Advanced
Subject: Chemistry
Chapter: States of Matter

JEE Advanced · Chemistry · States of Matter

One mole of triatomic vapours of an unknown substance effuses...

One mole of triatomic vapours of an unknown substance effuses

\frac{4}{3}

time faster than 1 mole of

\mathrm{O}_2

under identical conditions. If the density of unknown vapours at pressure

\mathbf{P}

and temperature

T

d

, which of the following statement(s) is/are correct for the unknown substance?

Options

ADensity at NTP = $0.8035 \mathrm{~g} / \mathrm{L}$
B $z$ (compressibility factor) $=\frac{15 \mathrm{P}}{4 \mathrm{dRT}}$
CVapour density = 9
D $z$ (compressibility factor) $=\frac{18 \mathrm{P}}{\mathrm{dRT}}$

Correct answer

D. $z$ (compressibility factor) $=\frac{18 \mathrm{P}}{\mathrm{dRT}}$

Step-by-step solution

\begin{aligned} & \frac{R_{\text {vap }}}{R_{\mathrm{O}_2}}=\sqrt{\frac{32}{M_{\text {vap }}}}=\frac{4}{3} \\ & \Rightarrow M_{\text {vap }}=\frac{32 \times 9}{16}=18 \\ & \Rightarrow \mathrm{VD}=9 ; \text { Density at NTP } \\ & =\frac{18}{22.4}=0.8035 \mathrm{gm} / \mathrm{L} \\ & Z=\frac{P V}{n R T}=\frac{P M}{d R T}=\frac{18 \mathrm{P}}{d R T}\end{aligned}

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