C_{v} and C_{p} denote the molar specific hea | vedclass

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(A) For a monoatomic ideal gas:$C_{v} = \frac{3}{2}R$,$C_{p} = \frac{5}{2}R$.Thus,$C_{p} - C_{v} = R$,$C_{p} + C_{v} = 4R$,$C_{p}/C_{v} = 5/3 \approx

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$(B, D)$

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(A) For a monoatomic ideal gas:$C_{v} = \frac{3}{2}R$,$C_{p} = \frac{5}{2}R$.Thus,$C_{p} - C_{v} = R$,$C_{p} + C_{v} = 4R$,$C_{p}/C_{v} = 5/3 \approx 1.67$,and $C_{p} \cdot C_{v} = 3.75 R^2$.For a diatomic ideal gas:$C_{v} = \frac{5}{2}R$,$C_{p} = \frac{7}{2}R$.Thus,$C_{p} - C_{v} = R$,$C_{p} + C_{v} = 6R$,$C_{p}/C_{v} = 7/5 = 1.4$,and $C_{p} \cdot C_{v} = 8.75 R^2$.Comparing the values:$1$. $C_{p} - C_{v} = R$ for both,so $(A)$ is incorrect.$2$. $C_{p} + C_{v}$ is $6R$ (diatomic) > $4R$ (monoatomic),so $(B)$ is correct.$3$. $C_{p}/C_{v}$ is $1.4$ (diatomic) $4$. $C_{p} \cdot C_{v}$ is $8.75 R^2$ (diatomic) > $3.75 R^2$ (monoatomic),so $(D)$ is correct.Therefore,the correct options are $(B)$ and $(D)$.

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