One mole of ideal monoatomic gas \([\gamma=(5 / 3)]\) is mixed with one mole of diatomic gas \([\gamma=(7 / 5)]\). What is \(\gamma\) for the nixture? \(\gamma\) denotes the ratio of specific heat at constant oressure to that at constant volume. (A) \((35 / 23)\) (B) \((23 / 15)\) (C) \((3 / 2)\) (D) \((4 / 3)\)

The specific heat ratio (γ) is defined as the ratio of the specific heat at constant pressure (\(C_p\)) to that at constant volume (\(C_v\)). We have the formula: $$ \gamma = \frac{C_p}{C_v} $$

For an ideal monoatomic gas, the degrees of freedom (\(f\)) is 3. The specific heat at constant volume (\(C_v\)) for a monoatomic gas can be calculated using the formula: $$ C_{v,mono} = \frac{f}{2}R $$ Where \(R\) is the universal gas constant. For monoatomic gas, we get: $$ C_{v,mono} = \frac{3}{2}R $$ From the given information, we know the specific heat ratio (\(\gamma_{mono}\)) for monoatomic gas is \(\frac{5}{3}\). We can use this to calculate the specific heat at constant pressure (\(C_{p,mono}\)) using the formula: $$ \gamma_{mono} = \frac{C_{p,mono}}{C_{v,mono}} \implies C_{p,mono} = \gamma_{mono} \times C_{v,mono} $$ By plugging in the values, we get: $$ C_{p,mono} = \frac{5}{3} \times \frac{3}{2}R = \frac{5}{2}R $$

For an ideal diatomic gas, the degrees of freedom (\(f\)) is 5. The specific heat at constant volume (\(C_v\)) for a diatomic gas can be calculated using the formula: $$ C_{v,di} = \frac{f}{2}R $$ For diatomic gas, we get: $$ C_{v,di} = \frac{5}{2}R $$ From the given information, we know the specific heat ratio (\(\gamma_{di}\)) for diatomic gas is \(\frac{7}{5}\). We can use this to calculate the specific heat at constant pressure (\(C_{p,di}\)) using the formula: $$ \gamma_{di} = \frac{C_{p,di}}{C_{v,di}} \implies C_{p,di} = \gamma_{di} \times C_{v,di} $$ By plugging in the values, we get: $$ C_{p,di} = \frac{7}{5} \times \frac{5}{2}R = \frac{7}{2}R $$

Since we have one mole of each type of gas, we can find the specific heat at constant volume and pressure for the mixture by taking the average of the specific heats of the two gasses. $$ C_{v,mixture} = \frac{C_{v,mono} + C_{v,di}}{2} = \frac{\frac{3}{2}R + \frac{5}{2}R}{2} = \frac{4}{2}R = 2R $$ $$ C_{p,mixture} = \frac{C_{p,mono} + C_{p,di}}{2} = \frac{\frac{5}{2}R + \frac{7}{2}R}{2} = \frac{6}{2}R = 3R $$

Now, we can calculate the specific heat ratio (\(\gamma_{mixture}\)) for the mixture by using the formula: $$ \gamma_{mixture} = \frac{C_{p,mixture}}{C_{v,mixture}} $$ By plugging in the values, we get: $$ \gamma_{mixture} = \frac{3R}{2R} = \frac{3}{2} $$ The specific heat ratio for the mixture is \(\frac{3}{2}\), so the correct answer is (C) \((\frac{3}{2})\).