Question

A mixture of one mole of a monoatomic gas and one mole of a diatomic gas (rigid) are kept at room temperature (\(27^\circ \text{C}\)). The ratio of specific heat of gases at constant volume respectively is:

• A. \(\frac{7}{5}\)
• B. \(\frac{3}{2}\)
• C. \(\frac{3}{5}\)
• D. \(\frac{5}{3}\)

Answer 1

The Correct Option is C

To determine the ratio of specific heats at constant volume for a gas mixture, the individual specific heats must first be computed.

Fundamental concepts include:

• For a monatomic gas, the degree of freedom is \( f = 3 \). The specific heat at constant volume \( C_V \) is thus: \(C_{V_\text{monoatomic}} = \frac{f}{2}R = \frac{3}{2}R\).
• For a rigid diatomic gas, the degree of freedom is \( f = 5 \). The specific heat at constant volume \( C_V \) is: \(C_{V_\text{diatomic}} = \frac{f}{2}R = \frac{5}{2}R\).

Considering one mole of a monatomic gas and one mole of a diatomic gas, the total number of moles is \( n = 2 \).

The average specific heat at constant volume for the mixture, \( C_{V_{\text{mix}}} \), is calculated as a weighted average:

\(C_{V_{\text{mix}}} = \frac{(1 \cdot C_{V_\text{monoatomic}}) + (1 \cdot C_{V_\text{diatomic}})}{2}\)

Upon substitution of the values:

\(C_{V_{\text{mix}}} = \frac{(1 \cdot \frac{3}{2}R) + (1 \cdot \frac{5}{2}R)}{2}\) \(C_{V_{\text{mix}}} = \frac{\frac{3}{2}R + \frac{5}{2}R}{2}\) \(C_{V_{\text{mix}}} = \frac{8}{4}R = 2R\)

The ratio of specific heats for the gases is then calculated as:

\(\text{Ratio} = \frac{C_{V_\text{monoatomic}}}{C_{V_\text{mix}}} = \frac{\frac{3}{2}R}{2R}\) \(\text{Ratio} = \frac{3}{4}\)

Therefore, the expected correct option was \(\frac{3}{4}\).

However, the provided correct answer is \(\frac{3}{5}\), which contradicts our calculated result.