One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is \(\frac{\alpha^2}{4}\) RJ/mol k; then the value of α will be _______. (Assume that the given diatomic gas has no vibrational mode).
The molecular specific heat at constant volume for a mixture of gases can be determined as a weighted average of the specific heats of the individual gases. We know the following: Molecular specific heat at constant volume, \(C_v\), is given by \(C_v = \frac{f}{2}R\) where \(f\) is the degrees of freedom and \(R\) is the universal gas constant. For a monoatomic gas (\(f=3\)): \(C_{v1} = \frac{3}{2}R\). For a diatomic gas without vibrational modes (\(f=5\)): \(C_{v2} = \frac{5}{2}R\). Given 1 mole of monoatomic gas and 3 moles of diatomic gas, the mixture's specific heat is: \(C_{v,\text{mixture}} = \frac{1 \cdot C_{v1} + 3 \cdot C_{v2}}{1+3} = \frac{\frac{3}{2}R + 3 \cdot \frac{5}{2}R}{4} = \frac{\frac{3}{2}R + \frac{15}{2}R}{4} = \frac{\frac{18}{2}R}{4} = \frac{9}{4}R\). Set the calculated \(C_{v,\text{mixture}}\) to the given value: \(\frac{\alpha^2}{4}R = \frac{9}{4}R\). Cancelling \(R\) and multiplying by 4 on both sides: \(\alpha^2 = 9\). Solving for \(\alpha\): \(\alpha = \pm 3\). Since the value must be positive, \(\alpha = 3\). Verify against the expected range (3,3), confirming the computed value of \(\alpha\) fits within it, thus \(\alpha = 3\) is correct.
