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 find the ratio of the specific heats at constant volume for a mixture of gases, we need to calculate the individual specific heats first.

Let's start with some fundamental concepts:

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

Given one mole of a monoatomic gas and one mole of a diatomic gas, the total number of moles \( n = 2 \).

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

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

Substituting the values:

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

Now let's find the ratio of the specific heats for the gases:

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

Thus, the correct option was expected to be \(\frac{3}{4}\).

However, the given correct answer is \(\frac{3}{5}\), which seems incorrect based on our calculation.

Answer 2

The specific heat capacities at constant volume are:
\[ (C_V)_{\text{mono}} = \frac{3}{2}R, \quad (C_V)_{\text{dia}} = \frac{5}{2}R. \]
The ratio is:
\[ \frac{(C_V)_{\text{mono}}}{(C_V)_{\text{dia}}} = \frac{\frac{3}{2}R}{\frac{5}{2}R} = \frac{3}{5}. \]
Final Answer: \(3 : 5\).