The heat absorbed by a mole of monoatomic gas at constant pressure, when its temperature is raised from 273 K to 373 K is Q. The heat absorbed by a mole of diatomic gas at constant volume, when its temperature is raised from 273 K to 323 K is $Q' .$ The ratio of Q to $Q'$ is

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Detailed Solution

For constant pressure, $Q = n C_{p} Δ T$

$Here n = 1, C_{p} (Mono) = (\frac{f}{2} + 1) R = (\frac{3}{2} + 1) R = \frac{5 R}{2}$

$Δ T = (373 K - 273 K) = 100 K$

$\Rightarrow Q = 1 (\frac{5 R}{2}) (100 K)$

For constant volume, $Q^{'} = n C_{\nabla} Δ T$

$Here n = 1, C_{\nabla} diatomic = \frac{f R}{2} = \frac{5 R}{2}$

$Δ T = (323 K - 273 K) = 50 K$

$\Rightarrow Q^{'} = (1) \cdot (\frac{5 R}{2}) (50 K) \Rightarrow \frac{Q}{Q^{'}} = \frac{100 K}{50 K} = 2$

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