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(D) For an ideal gas,the internal energy is given by $U = U_0 + nC_vT$. Given the relation $U = 2 + 2PV$. Using the ideal gas equation $PV = nRT$,we s

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a mixture of monoatomic and diatomic gases

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(D) For an ideal gas,the internal energy is given by $U = U_0 + nC_vT$. Given the relation $U = 2 + 2PV$. Using the ideal gas equation $PV = nRT$,we substitute this into the given relation: $U = 2 + 2nRT$. Differentiating with respect to temperature $T$,we get: $\frac{dU}{dT} = 2nR$. Since $C_v = \frac{1}{n} \frac{dU}{dT}$,we have $C_v = 2R$. For a monoatomic gas,$C_v = \frac{3}{2}R = 1.5R$. For a diatomic gas,$C_v = \frac{5}{2}R = 2.5R$. Since the calculated $C_v = 2R$ lies between $1.5R$ and $2.5R$,the gas must be a mixture of monoatomic and diatomic gases.

The relation between internal energy $U$,pressure $P$,and volume $V$ for an ideal gas is given by $U = 2 + 2PV$. The gas is:

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