One mole of a monoatomic gas \([\gamma=(5 / 3)]\) is mixed with one mole of A diatomic gas \([\gamma=(7 / 5)]\) what will be value of \(\gamma\) for mixture ? (A) \(1.454\) (B) \(1.4\) (C) \(1.54\) (D) \(1.5\)

For both the monoatomic and diatomic gas, we will calculate their heat capacities \(C_P\) and \(C_V\). Using the given \(\gamma\) values and the relation between the heat capacities \(C_P = C_V + R\). First, we have: Monoatomic gas - \(\gamma_1 = \frac{5}{3}\). Solve for \(C_{V1}\), the heat capacity at constant volume: \(C_{V1} = R \frac{\gamma_1 - 1}{\gamma_1 - 1} = R\) Now calculate \(C_{P1}\), the heat capacity at constant pressure: \(C_{P1} = C_{V1} + R = R + R = 2R\) Diatomic gas - \(\gamma_2 = \frac{7}{5}\). Solve for \(C_{V2}\), the heat capacity at constant volume: \(C_{V2} = R \frac{\gamma_2 - 1}{\gamma_2 - 1} = \frac{5}{2} R\) Now calculate \(C_{P2}\), the heat capacity at constant pressure: \(C_{P2} = C_{V2} + R = \frac{5}{2} R + R = \frac{7}{2} R\)

Since both gases have one mole, the average heat capacities can be simply calculated by taking the average of the individual heat capacities. \(C_{V_{mix}} = \frac{C_{V1}+C_{V2}}{2} = \frac{1}{2}(R + \frac{5}{2} R) = \frac{7}{4} R\) And, \(C_{P_{mix}} = \frac{C_{P1}+C_{P2}}{2} = \frac{1}{2}(2R + \frac{7}{2} R) = \frac{11}{4} R\).

Now, we can calculate the value of \(\gamma\) for the mixture using the formula: \(\gamma_{mix} = \frac{C_{P_{mix}}}{C_{V_{mix}}} = \frac{\frac{11}{4} R}{\frac{7}{4} R} = \frac{11}{7}\). Now convert \(\frac{11}{7}\) to a decimal value: \(\frac{11}{7} = 1.5714\) Comparing the obtained value with the given choices in the problem, we can see that option (C) is the closest. So, the value of \(\gamma\) for the mixture is approximately \(\boxed{1.54}\).