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(A, C) For a diatomic gas, the degree of freedom $f = 5$. The internal energy is $U = \frac{n f R T}{2} = \frac{5 R}{2} (\alpha t + \frac{1}{2} \beta

Vedclass · Accepted Answer

the rate of increase in internal energy is $\frac{5}{2} R(\alpha+\beta t)$

Vedclass · Answer

(A, C) For a diatomic gas, the degree of freedom $f = 5$. The internal energy is $U = \frac{n f R T}{2} = \frac{5 R}{2} (\alpha t + \frac{1}{2} \beta t^2)$.The rate of increase in internal energy is $\frac{dU}{dt} = \frac{5 R}{2} (\alpha + \beta t)$.Since the pressure is constant, the heat supplied is $dQ = n C_p dT$. For a diatomic gas, $C_p = \frac{7}{2} R$.Thus, the power supplied by the heating element is $i^2 r = \frac{dQ}{dt} = n C_p \frac{dT}{dt} = 1 	imes \frac{7}{2} R 	imes (\alpha + \beta t)$.Therefore, the current $i = \sqrt{\frac{7 R}{2 r} (\alpha + \beta t)}$.From the ideal gas law $PV = nRT$, since $P$ is constant, $V = \frac{nRT}{P} = \frac{R}{P} (\alpha t + \frac{1}{2} \beta t^2)$.The position of the piston $x = \frac{V}{A} = \frac{R}{PA} (\alpha t + \frac{1}{2} \beta t^2)$.The velocity $v = \frac{dx}{dt} = \frac{R}{PA} (\alpha + \beta t)$.The acceleration $a = \frac{dv}{dt} = \frac{R \beta}{PA}$, which is constant. Thus, options $(a)$ and $(c)$ are correct.

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