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(D) The number of moles for each gas are: $n_1 = 4$ (hydrogen,diatomic),$n_2 = 2$ (helium,monatomic),$n_3 = 1$ (water vapour,polyatomic).The molar spe

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$\frac{23}{7} R$

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(D) The number of moles for each gas are: $n_1 = 4$ (hydrogen,diatomic),$n_2 = 2$ (helium,monatomic),$n_3 = 1$ (water vapour,polyatomic).The molar specific heat at constant volume $(C_v)$ for each gas is:For hydrogen (diatomic): $(C_v)_1 = \frac{5}{2} R$For helium (monatomic): $(C_v)_2 = \frac{3}{2} R$For water vapour (polyatomic): $(C_v)_3 = 3 R$The equivalent molar specific heat at constant volume for the mixture is given by:$(C_v)_{	ext{mix}} = \frac{n_1(C_v)_1 + n_2(C_v)_2 + n_3(C_v)_3}{n_1 + n_2 + n_3}$$(C_v)_{	ext{mix}} = \frac{4 	imes \frac{5}{2} R + 2 	imes \frac{3}{2} R + 1 	imes 3 R}{4 + 2 + 1} = \frac{10 R + 3 R + 3 R}{7} = \frac{16}{7} R$Using the relation $C_p = C_v + R$ for the mixture:$(C_p)_{	ext{mix}} = (C_v)_{	ext{mix}} + R = \frac{16}{7} R + R = \frac{23}{7} R$.

Four moles of hydrogen,two moles of helium,and one mole of water vapour form an ideal gas mixture. What is the molar specific heat at constant pressure of the mixture?

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