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(D) For a diatomic gas,the degrees of freedom $f = 5$.At constant pressure,the heat supplied is $\Delta Q = n C_p \Delta T$.The change in internal ene

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$7: 5: 2$

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(D) For a diatomic gas,the degrees of freedom $f = 5$.At constant pressure,the heat supplied is $\Delta Q = n C_p \Delta T$.The change in internal energy is $\Delta U = n C_v \Delta T$.The work done is $\Delta W = n R \Delta T$.Using the relation $C_p = \frac{f+2}{2} R$ and $C_v = \frac{f}{2} R$,we have:$\Delta Q = n \left( \frac{f+2}{2} \right) R \Delta T = n \left( \frac{5+2}{2} \right) R \Delta T = \frac{7}{2} n R \Delta T$.$\Delta U = n \left( \frac{f}{2} \right) R \Delta T = n \left( \frac{5}{2} \right) R \Delta T = \frac{5}{2} n R \Delta T$.$\Delta W = n R \Delta T = \frac{2}{2} n R \Delta T$.Thus,the ratio $\Delta Q : \Delta U : \Delta W = \frac{7}{2} : \frac{5}{2} : \frac{2}{2} = 7 : 5 : 2$.

Heat is supplied to a diatomic gas at constant pressure. The ratio of $\Delta Q: \Delta U: \Delta W$ is
[Given $\rightarrow \Delta Q=$ heat supplied,$\Delta U=$ change in internal energy,$\Delta W$ $=$ work done]

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Heat is supplied to a diatomic gas at constant pressure. The ratio of $\Delta Q: \Delta U: \Delta W$ is[Given $\rightarrow \Delta Q=$ heat supplied,$\Delta U=$ change in internal energy,$\Delta W$ $=$ work done]