Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is :
- \(\frac{9}{4}R\)
- \(\frac{7}{4}R\)
- \(\frac{3}{2}R\)
- \(\frac{5}{2}R\)
The Correct Option is A
Approach Solution - 1
For a monatomic gas, the molar specific heat at constant volume is \(C_V = \frac{3}{2}R\), and for a diatomic gas, \(C_V = \frac{5}{2}R\). The total heat capacity \(C_V\) of the mixture is given by the weighted average formula:
\[ C_V = \frac{n_1C_{V1} + n_2C_{V2}}{n_1 + n_2} \]
Where:
- \(n_1 = 2\) is the number of moles of the monatomic gas.
- \(n_2 = 6\) is the number of moles of the diatomic gas.
- \(C_{V1} = \frac{3}{2}R\) is the molar specific heat of the monatomic gas.
- \(C_{V2} = \frac{5}{2}R\) is the molar specific heat of the diatomic gas.
Substituting the values into the equation:
\[ C_V = \frac{2 \times \frac{3}{2}R + 6 \times \frac{5}{2}R}{2 + 6} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R \]
Thus, the molar specific heat of the mixture at constant volume is \(\frac{9}{4}R\).
Approach Solution -2
The problem asks to determine the molar specific heat of a mixture at constant volume, which is formed by mixing 2 moles of a monoatomic gas with 6 moles of a diatomic gas.
Concept Used:
The molar specific heat of a gas at constant volume (\( C_v \)) is related to its degrees of freedom (\(f\)) by the formula derived from the equipartition of energy theorem:
\[ C_v = \frac{f}{2} R \]
where \( R \) is the universal gas constant.
1. For a monoatomic gas: There are 3 translational degrees of freedom. So, \( f_1 = 3 \). \[ C_{v1} = \frac{3}{2} R \]
2. For a diatomic gas: At normal temperatures, there are 3 translational and 2 rotational degrees of freedom (vibrational modes are not considered). So, \( f_2 = 5 \). \[ C_{v2} = \frac{5}{2} R \]
3. For a mixture of gases: The molar specific heat at constant volume for the mixture (\( C_{v, \text{mix}} \)) is the weighted average of the molar specific heats of the individual gases, based on their mole fractions. The formula is: \[ C_{v, \text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \] where \( n_1 \) and \( n_2 \) are the number of moles of the respective gases.
Step-by-Step Solution:
Step 1: Identify the given quantities.
Number of moles of the monoatomic gas, \( n_1 = 2 \) moles.
Number of moles of the diatomic gas, \( n_2 = 6 \) moles.
Step 2: Determine the molar specific heat at constant volume for the monoatomic gas (\( C_{v1} \)).
For a monoatomic gas, the number of degrees of freedom is \( f_1 = 3 \).
\[ C_{v1} = \frac{f_1}{2} R = \frac{3}{2} R \]
Step 3: Determine the molar specific heat at constant volume for the diatomic gas (\( C_{v2} \)).
For a diatomic gas, the number of degrees of freedom is \( f_2 = 5 \).
\[ C_{v2} = \frac{f_2}{2} R = \frac{5}{2} R \]
Step 4: Use the formula for the molar specific heat of a mixture at constant volume.
\[ C_{v, \text{mix}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} \]
Step 5: Substitute the known values into the formula and perform the calculation.
\[ C_{v, \text{mix}} = \frac{(2) \left( \frac{3}{2} R \right) + (6) \left( \frac{5}{2} R \right)}{2 + 6} \]
Simplify the numerator:
\[ \text{Numerator} = 3R + (3)(5R) = 3R + 15R = 18R \]
Simplify the denominator:
\[ \text{Denominator} = 8 \]
Now, compute the final value:
\[ C_{v, \text{mix}} = \frac{18R}{8} = \frac{9R}{4} \]
The molar specific heat of the mixture at constant volume is \(\frac{9}{4} R\).
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