JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2020
- Medium
- Numerical
To raise the temperature of a certain mass of gas by \(50^{\circ} C\) at a constant pressure, \(160\) calories of heat is required. When the same mass of gas is cooled by \(100^{\circ} C\) at constant volume, \(240\) calories of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal) \(?\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\(nC _{ P }(50)=160\)
\(nC _{ v }(100)=240\)
\(\Rightarrow \frac{ C _{ p }}{2 C _{ v }}=\frac{160}{240}=\frac{\gamma}{2}\)
\(\therefore \gamma=\frac{4}{3}\) and \(f =\frac{2}{\gamma-1}=6\)
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