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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) \(?\)

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Answer & Solution

Correct Answer

(C) \(6\)

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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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