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JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2013
- Medium
- MCQ
Figure shows the variation in temperature \(\left( {\Delta T} \right)\) with the amount of heat supplied \((Q)\) in an isobaric process corresponding to a monoatomic \((M)\), diatomic \((D)\) and a polyatomic \((P)\) gas. The initial state of all the gases are the same and the scales for the two axes coincide. Ignoring vibrational degrees of freedom, the lines \(a, b\) and \(c\) respectively correspond to
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(M, D\) and \(P\)
Step-by-step Solution
Detailed explanation
On giving same amount of heat at constant pressure, there is no change in temperature for mono, dia and polyatomic.
\((\Delta \mathrm{Q})_{\mathrm{P}}=\mu \mathrm{C}_{\mathrm{p}} \Delta \mathrm{T}\left(\mu=\begin{array}{l}{\text { No. of molecules }} \\ {\text { Avogedro's na }}\end{array}\right)\)
or \(\quad \Delta \mathrm{T} \propto \frac{1}{\text { no of molecules }}\)
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