If V=1 litre, 10 mole of H_2 and 10 mole of N_2 gas are mixed at temperature 26^ C , then calculate pressure of the gas.

Correct answer: A. 491 atm According to ideal gas equation, pV=nRT . Given, V=1 ~L, n_1=10 mole, n_2=10 mole. T=26^ C=299 ~K . On substituting the given values in above equation, we get. aligned p & = 20 ~mol × 0.0821 ~L ~atm ~mol^ -1 ~K^ -1 × 299 ~K 1 ~L & =491 ~atm . aligned

JIPMER 2019

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

JIPMER 2019 · Chemistry · States of Matter

If V=1 litre, 10 mole of H_2 and 10 mole of N_2 gas are mixed at...

\mathrm{V}=1

litre, 10 mole of

\mathrm{H}_2

and 10 mole of

\mathrm{N}_2

gas are mixed at temperature

26^{\circ} \mathrm{C}

, then calculate pressure of the gas.

Options

A491 atm
B300 atm
C550 atm
D600 atm

Correct answer

A. 491 atm

Step-by-step solution

According to ideal gas equation,

\mathrm{pV}=\mathrm{nRT}

. Given,

\mathrm{V}=1 \mathrm{~L}, \mathrm{n}_1=10

mole,

\mathrm{n}_2=10

mole.

\mathrm{T}=26^{\circ} \mathrm{C}=299 \mathrm{~K} \text{. }

On substituting the given values in above equation, we get.

\begin{aligned} \mathrm{p} & =\frac{20 \mathrm{~mol} \times 0.0821 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 299 \mathrm{~K}}{1 \mathrm{~L}} \\ & =491 \mathrm{~atm} . \end{aligned}

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