The volume occupied by 1.8 ~g of water vapour at 374^ C and 1 bar pressure will be [Use R=0.083 bar L K^ -1 ~mol^ -1 ]

Correct answer: D. 5.37 ~L According to ideal gas equation, aligned & pV=nRT & V= nRT p = w M.wt RT p (i)[ n= w M · wt ] aligned Given, w=1.8 ~g, ~T=374^ ^ C =(374+273) K= 647 ~K p= 1 bar, R= 0.083 bar LK^ -1 ~mol^ -1 On substituting the given values in Eq. (i), we get aligned V& = 1.8 ~g 18 ~g ~mol^ -1 × 0.083 bar LK^ -1 ~mol^ -1 × 647 ~K 1 bar & =5.37 ~L aligned

NEET 2019

Exam: NEET
Year: 2019
Subject: Chemistry
Chapter: States of Matter

NEET 2019 · Chemistry · States of Matter

The volume occupied by 1.8 ~g of water vapour at 374^ C and 1 bar...

The volume occupied by

1.8 \mathrm{~g}

of water vapour at

374^{\circ} \mathrm{C}

and 1 bar pressure will be [Use

\mathrm{R}=0.083

bar

\mathrm{L} \mathrm{K}^{-1} \mathrm{~mol}^{-1}

]

Options

A $96.66 \mathrm{~L}$
B $55.87 \mathrm{~L}$
C $3.10 \mathrm{~L}$
D $5.37 \mathrm{~L}$

Correct answer

D. $5.37 \mathrm{~L}$

Step-by-step solution

According to ideal gas equation,

\begin{aligned}& \mathrm{pV}=\mathrm{nRT}\\ & \mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{p}}=\frac{\mathrm{w}}{\text{ M.wt }} \frac{\mathrm{RT}}{\mathrm{p}} \ldots(\mathrm{i})\left[\because \mathrm{n}=\frac{\mathrm{w}}{\mathrm{M} \cdot \mathrm{wt}}\right] \end{aligned}

Given,

\mathrm{w}=1.8 \mathrm{~g}, \mathrm{~T}=374^{{}^{\circ}} \mathrm{C}

=(374+273)

\mathrm{K}=

647

\mathrm{~K} \mathrm{p}=

\text{ bar, } \mathrm{R}=

0.083

\mathrm{bar} \mathrm{LK}^{-1} \mathrm{~mol}^{-1}

On substituting the given values in Eq. (i), we get

\begin{aligned}\mathrm{V}& =\frac{1.8 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}} \times \frac{0.083 \mathrm{bar} \mathrm{LK}^{-1} \mathrm{~mol}^{-1} \times 647 \mathrm{~K}}{1 \mathrm{bar}}\\ & =5.37 \mathrm{~L} \end{aligned}

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