\(\text{If three moles of monoatomic gas } \left( \gamma = \frac{5}{3} \right) \text{ is mixed with two moles of a diatomic gas } \left( \gamma = \frac{7}{5} \right),\)\(\text{the value of adiabatic exponent } \gamma \text{ for the mixture is:}\)
- 1.75
- 1.4
- 1.52
- 1.35
The Correct Option is C
Approach Solution - 1
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 γmixture 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:
\[ \gamma_{\text{mixture}} = \frac{29}{19} \approx 1.52 \]
Approach Solution -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 \( \frac{29}{19} = 1.52 \).
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