Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom.

Question:

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of $\mathrm{T}$. The total internal energy, U of a mole of this gas, and

the value of $\gamma\left(=\frac{C_{\mathrm{P}}}{C_{\mathrm{v}}}\right)$ given, respectively, by:

 

  1. $\mathrm{U}=\frac{5}{2} \mathrm{RT}$ and $\gamma=\frac{6}{5}$

  2. $\mathrm{U}=5 \mathrm{RT}$ and $\gamma=\frac{7}{5}$

  3. $\mathrm{U}=5 \mathrm{RT}$ and $\gamma=\frac{6}{5}$

  4. $\mathrm{U}=\frac{5}{2} \mathrm{RT}$ and $\gamma=\frac{7}{5}$


Correct Option: , 4

Solution:

Total degree of freedom $=3+2=5$

$\mathrm{U}=\frac{\mathrm{nfRT}}{2} \Rightarrow \frac{5 \mathrm{RT}}{2}$

$\gamma \Rightarrow \frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}} \Rightarrow 1+\frac{2}{\mathrm{f}} \Rightarrow 1+\frac{2}{5} \Rightarrow \frac{7}{5}$

Ans. (4)

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now