JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
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 \(\gamma\left(=\frac{ C _{ P }}{ C _{ v }}\right)\) given, respectively, by
- A \(U =\frac{5}{2} RT\) and \(\gamma=\frac{6}{5}\)
- B \(U =5 RT\) and \(\gamma=\frac{7}{5}\)
- C \(U =5 RT\) and \(\gamma=\frac{6}{5}\)
- D \(U =\frac{5}{2} RT\) and \(\gamma=\frac{7}{5}\)
Answer & Solution
Correct Answer
(D) \(U =\frac{5}{2} RT\) and \(\gamma=\frac{7}{5}\)
Step-by-step Solution
Detailed explanation
Total degree of freedom \(=3+2=5\) \(U =\frac{ nfRT }{2} \Rightarrow \frac{5 RT }{2}\) \(\gamma \Rightarrow \frac{ C _{ P }}{ C _{ V }} \Rightarrow 1+\frac{2}{ f } \Rightarrow 1+\frac{2}{5} \Rightarrow \frac{7}{5}\)
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