One mole of diatomic gas having rotational mo | vedclass

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Question

(B) For a diatomic gas with rotational modes only, the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). Thus, $C_v = \frac{f}{2}R = \f

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) For a diatomic gas with rotational modes only, the degrees of freedom $f = 5$ ($3$ translational + $2$ rotational). Thus, $C_v = \frac{f}{2}R = \frac{5}{2}R$.Change in internal energy $\Delta U = nC_v \Delta T = 1 	imes \frac{5}{2} 	imes 8.3 	imes 1.2 = 24.9 	ext{ J}$.Work done by the gas $W = P \Delta V = P(A \Delta x)$.Given $P = 10^5 	ext{ Pa}$, $A = 4 	imes 10^{-4} 	ext{ m}^2$, and $\Delta x = 25 	imes 10^{-3} 	ext{ m}$.$W = 10^5 	imes 4 	imes 10^{-4} 	imes 25 	imes 10^{-3} = 1 	ext{ J}$.Total heat supplied $Q = \Delta U + W = 24.9 + 1 = 25.9 	ext{ J}$.Comparing with the given options, $25 	ext{ J}$ is the closest value.

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