Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let    be the ratio of molar specific heat at constant pressure

Let  \mathrm{\gamma _{1}}  be the ratio of molar specific heat at constant pressure and molar specific heat at constant volume of a monoatomic gas and \mathrm{\gamma _{2}} be the similar ratio of diatomic gas. Considering the diatomic gas molecule as a rigid rotator, the ratio,\mathrm{\frac{\gamma _{1}}{\gamma _{2}}} is :
 

Option: 1

\frac{25}{21}


Option: 2

\frac{35}{27}


Option: 3

\frac{21}{25}


Option: 4

\frac{27}{35}


Answers (1)

For monoatomic gas
 
\gamma_1=\frac{5}{3}
 
For diatomic gas at low temperatures
 \gamma_2=\frac{7}{5} \\
\begin{aligned} & \therefore \frac{\gamma_1}{\gamma_2}=\frac{\frac{5}{3}}{\frac{7}{5}}=\frac{25}{21} \end{aligned}

Posted by

Sumit Saini

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

Amrita BTech Admission 2025: JEE Main-based registration deadline extended to August 30

Latest Question