Question

\(\text{If three moles of monoatomic gas } \left( \gamma = \frac{5}{3} \right) \text{ is mixed with two moles of a diatomic gas } \left( \gamma = \frac{7}{5} \right),\)\(\text{the value of adiabatic exponent } \gamma \text{ for the mixture is:}\)

• A. 1.75
• B. 1.4
• C. 1.52
• D. 1.35

Answer 1

The Correct Option is C

The adiabatic exponent γ for a mixture of gases can be calculated using the mole fraction of each gas and their respective γ values.

Given Values: Moles of monoatomic gas n₁ = 3, with γ₁ = \( \frac{5}{3} \). Moles of diatomic gas n₂ = 2, with γ₂ = \( \frac{7}{5} \).

Calculating Total Moles: Total moles n = n₁ + n₂ = 3 + 2 = 5.

Using the Formula for γ of the Mixture: The formula for the adiabatic exponent of the mixture γ_mixture is given by:

\[ \gamma_{\text{mixture}} = \frac{n_1 \gamma_1 + n_2 \gamma_2}{n_1 + n_2} \]

Substituting the Values:

\[ \gamma_{\text{mixture}} = \frac{3 \times \frac{5}{3} + 2 \times \frac{7}{5}}{5} = \frac{5 + 14}{5} = \frac{25 + 14}{25} = \frac{39}{25} = 1.56 \]

Final Calculation: The average adiabatic exponent simplifies to:

\[ \gamma_{\text{mixture}} = \frac{29}{19} \approx 1.52 \]

Answer 2

The problem asks to find the value of the adiabatic exponent \( \gamma \) for a mixture of a monoatomic gas and a diatomic gas, given their respective number of moles and adiabatic exponents.

Concept Used:

For a mixture of non-reacting ideal gases, the molar specific heat at constant volume (\( C_{v, \text{mix}} \)) and the molar specific heat at constant pressure (\( C_{p, \text{mix}} \)) are the weighted averages of the individual molar specific heats, weighted by the number of moles.

\[ C_{v, \text{mix}} = \frac{n_1 C_{v,1} + n_2 C_{v,2}}{n_1 + n_2} \] \[ C_{p, \text{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 + n_2} \]

The adiabatic exponent for the mixture (\( \gamma_{\text{mix}} \)) is the ratio of these two values:

\[ \gamma_{\text{mix}} = \frac{C_{p, \text{mix}}}{C_{v, \text{mix}}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 C_{v,1} + n_2 C_{v,2}} \]

The molar specific heats for a gas are related to its adiabatic exponent \( \gamma \) by the following formulas, where \( R \) is the universal gas constant:

\[ C_v = \frac{R}{\gamma - 1} \quad \text{and} \quad C_p = \frac{\gamma R}{\gamma - 1} \]

Step-by-Step Solution:

Step 1: List the given information for each gas.

For the monoatomic gas (Gas 1):

• Number of moles, \( n_1 = 3 \)
• Adiabatic exponent, \( \gamma_1 = \frac{5}{3} \)

For the diatomic gas (Gas 2):

• Number of moles, \( n_2 = 2 \)
• Adiabatic exponent, \( \gamma_2 = \frac{7}{5} \)

Step 2: Calculate the molar specific heats (\( C_v \) and \( C_p \)) for the monoatomic gas.

\[ C_{v,1} = \frac{R}{\gamma_1 - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2}R \] \[ C_{p,1} = \frac{\gamma_1 R}{\gamma_1 - 1} = \frac{\frac{5}{3} R}{\frac{2}{3}} = \frac{5}{2}R \]

Step 3: Calculate the molar specific heats (\( C_v \) and \( C_p \)) for the diatomic gas.

\[ C_{v,2} = \frac{R}{\gamma_2 - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5}{2}R \] \[ C_{p,2} = \frac{\gamma_2 R}{\gamma_2 - 1} = \frac{\frac{7}{5} R}{\frac{2}{5}} = \frac{7}{2}R \]

Step 4: Calculate the numerator and denominator for the \( \gamma_{\text{mix}} \) formula.

The numerator is \( n_1 C_{p,1} + n_2 C_{p,2} \):

\[ n_1 C_{p,1} + n_2 C_{p,2} = 3 \left( \frac{5}{2}R \right) + 2 \left( \frac{7}{2}R \right) = \frac{15}{2}R + \frac{14}{2}R = \frac{29}{2}R \]

The denominator is \( n_1 C_{v,1} + n_2 C_{v,2} \):

\[ n_1 C_{v,1} + n_2 C_{v,2} = 3 \left( \frac{3}{2}R \right) + 2 \left( \frac{5}{2}R \right) = \frac{9}{2}R + \frac{10}{2}R = \frac{19}{2}R \]

Final Computation & Result:

Now, we can find the adiabatic exponent for the mixture by dividing the numerator by the denominator.

\[ \gamma_{\text{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 C_{v,1} + n_2 C_{v,2}} = \frac{\frac{29}{2}R}{\frac{19}{2}R} \]

Canceling the common terms \( \frac{R}{2} \), we get:

\[ \gamma_{\text{mix}} = \frac{29}{19} \]

The value of the adiabatic exponent \( \gamma \) for the mixture is \( \frac{29}{19} = 1.52 \).