Molecules of an ideal gas are known to have three translational degrees

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 $T$. The total internal energy, U of a mole of this gas, and the

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

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

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

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

  4. (4) $\mathrm{U}=5 \mathrm{RT}$ and $\gamma=\frac{6}{5}$

Correct Option: , 3

Solution:

(3) Total degree of freedom $f=3+2=5$

Total energy, $U=\frac{n f R T}{2}=\frac{5 R T}{2}$

And $\gamma=\frac{C_{p}}{C_{v}}=1+\frac{2}{f}=1+\frac{2}{5}=\frac{7}{5}$

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