The ratio of the work done,change in internal | vedclass

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(D) For a diatomic gas,the degrees of freedom $f = 5$. At constant pressure,the heat absorbed is given by $dQ = n C_p dT$. The work done is $dW = P dV

Vedclass · Accepted Answer

$2: 5: 7$

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(D) For a diatomic gas,the degrees of freedom $f = 5$. At constant pressure,the heat absorbed is given by $dQ = n C_p dT$. The work done is $dW = P dV = n R dT$. The change in internal energy is $dU = n C_v dT$. We know that $C_v = \frac{f}{2} R = \frac{5}{2} R$ and $C_p = C_v + R = \frac{7}{2} R$. Thus,the ratio $dW : dU : dQ$ is $nR dT : \frac{5}{2} nR dT : \frac{7}{2} nR dT$. Dividing by $nR dT$,we get $1 : \frac{5}{2} : \frac{7}{2}$. Multiplying by $2$,we get $2 : 5 : 7$.

The ratio of the work done,change in internal energy,and heat absorbed when a diatomic gas expands at constant pressure is

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