1 mole of a gas with \gamma =7/5 is mixed with 1 mole of a gas with \gamma =5/3 , then the value of \gamma  for the resulting mixture is

  • Option 1)

    7/5\;

  • Option 2)

    \; 2/5\;

  • Option 3)

    \; 24/16\;

  • Option 4)

    \; 12/7

 

Answers (1)

As we learnt in

First law in isobaric process -

\Delta U= n\, C_{v}\Delta T

= n\frac{R}{\gamma -1}\Delta T
 

- wherein

\Delta Q= \Delta U+W

         = n\frac{\gamma \: R}{\gamma -1}\cdot \Delta T

\Delta Q= nC_{p}\: \Delta T

 \gamma 1 =\frac{7}{5}\rightarrow CV_{1}= \frac{R}{\gamma 1-1}= \frac{5R}{2},C_{P1}=\frac{7R}{2}

\gamma 2 =\frac{5}{3}\rightarrow CV_{2}= \frac{R}{\gamma 2-1}= \frac{3R}{2},C_{P2}=\frac{5R}{2}

 \gamma^{mixture}=\frac{C_{p}^{mix}}{C_{v}^{mix}}=\frac{n_{1}C_{p1}+n_{2}C_{p2}}{n_{1}C_{v1}+n_{2}C_{v2}}= \frac{6R}{4R}=1.5


Option 1)

7/5\;

Option 2)

\; 2/5\;

Option 3)

\; 24/16\;

Option 4)

\; 12/7

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 6999/- ₹ 5/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Knockout JEE Main January 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions