Question:
Figure shows the variation in temperature (\(\Delta T\)) 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 
Figure shows the variation in temperature (\(\Delta T\)) 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
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In an isobaric process \(Q = nC_p\Delta T\), so the slope of the \(Q\)-\(\Delta T\) line is \(nC_p\); read the slopes of a, b, c from the figure.
Updated On: Oct 31, 2026
- P, M and D
- M, D and P
- P, D and M
- D, M and P
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The Correct Option is C
Solution and Explanation
For the same initial state, all gases have the same number of moles \(n\). At constant pressure \(Q = nC_p\Delta T\), so the slope \(\frac{Q}{\Delta T} = nC_p\).
Ignoring vibration: monoatomic \(C_p = \frac52R\); diatomic \(C_p = \frac72R\); polyatomic (non-linear, 6 degrees of freedom) \(C_p = 4R\).
In the figure \(Q\) is on the vertical axis and \(\Delta T\) on the horizontal axis; line \(a\) is steepest, then \(b\), then \(c\), so the slopes are \(a \gt b \gt c\).
Hence \(a\) = P (largest \(C_p\)), \(b\) = D, \(c\) = M.
Answer: (3) P, D and M
Ignoring vibration: monoatomic \(C_p = \frac52R\); diatomic \(C_p = \frac72R\); polyatomic (non-linear, 6 degrees of freedom) \(C_p = 4R\).
In the figure \(Q\) is on the vertical axis and \(\Delta T\) on the horizontal axis; line \(a\) is steepest, then \(b\), then \(c\), so the slopes are \(a \gt b \gt c\).
Hence \(a\) = P (largest \(C_p\)), \(b\) = D, \(c\) = M.
Answer: (3) P, D and M
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