Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A closed container of volume contains a mixture of Neon and Argon gases

A closed container of volume 0.02 \mathrm{~m}^{3} contains a mixture of Neon and Argon gases at a temperature of 27^{\circ} \mathrm{C} and pressure of 1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}.The total mass of the mixture is 28 \mathrm{gm}. If the gram molecular weight of Neon and Argon are 20 and 40 respectively, find the masses of the individual gasses in the container assuming them to be ideal. \mathrm{(R= 8.314 J/moles K.)}

Option: 1

2gm,12gm


Option: 2

3 gm,8gm


Option: 3

4 gm,24 gm


Option: 4

4 gm,12gm


Answers (1)

best_answer

Let masses of neon and Argon are \mathrm{m_{1}} and \mathrm{m_{2}}

\mathrm{then \, m_{1}+m_{2}=28 \quad \ldots(i)}

From perfect gas equation

\mathrm{\mathrm{PV}=\mathrm{nRT}}

\mathrm{=\left(\frac{m_{1}}{M_{1}}+\frac{m_{2}}{M_{2}}\right) R T=\left(\frac{m_{1}}{20}+\frac{m_{2}}{40}\right) R T}
\mathrm{i.e. \frac{m_{1}}{20}+\frac{m_{2}}{40}=\frac{P V}{R T}}
\mathrm{=\frac{1 \times 10^{5} \times 0.02}{8.314 \times 300}}
\mathrm{i.e. \quad 2 m_{1}+m_{2}=32} \quad \ldots(ii)

\therefore  from eq. (I) and (ii)
\mathrm{m}_{1}=4 \mathrm{gm}, \mathrm{m}_{2}=24 \mathrm{gm}

Posted by

Pankaj

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question