Question

2 moles of an ideal gas with \( \frac{C_p}{C_v} = \frac{5}{3} \) are mixed with 3 moles of another ideal gas with \( \frac{C_p}{C_v} = \frac{4}{3} \). The value of \( \frac{C_p}{C_v} \) for the mixture is

• A. 1.5
• B. 1.42
• C. 1.48
• D. 1.6

Hint:

Instead of calculating \( C_p \) and \( C_v \) separately, you can use the direct formula for the mixture ratio \( \gamma_{\text{mix}} \):
\[ \frac{n_1 + n_2}{\gamma_{\text{mix}} - 1} = \frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1} \]
Substituting the values gives:
\[ \frac{5}{\gamma_{\text{mix}} - 1} = \frac{2}{2/3} + \frac{3}{1/3} = 3 + 9 = 12 \implies \gamma_{\text{mix}} - 1 = \frac{5}{12} \implies \gamma_{\text{mix}} = \frac{17}{12} \approx 1.42 \]
This direct formula is much faster and less prone to calculation errors.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The ratio of specific heats \(\gamma = C_p / C_v\) is a fundamental property of an ideal gas.
When two ideal gases are mixed, the mixture's molar heat capacities (\(C_{v,mix}\) and \(C_{p,mix}\)) are the weighted averages of the individual molar heat capacities based on the number of moles.
By calculating the total internal energy change or using the mixing rule for heat capacities, we can determine the resulting \(\gamma\) of the mixture.
Step 2: Key Formula or Approach:
1. \(C_{v,i} = \frac{R}{\gamma_i - 1}\).
2. \(C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2}\).
3. \(\gamma_{mix} = 1 + \frac{R}{C_{v,mix}} = \frac{n_1 \gamma_1 / (\gamma_1 - 1) + n_2 \gamma_2 / (\gamma_2 - 1)}{n_1 / (\gamma_1 - 1) + n_2 / (\gamma_2 - 1)}\).
Step 3: Detailed Explanation:
Gas 1: \(n_1 = 2\), \(\gamma_1 = 5/3 \implies C_{v1} = \frac{R}{5/3 - 1} = \frac{R}{2/3} = 1.5R\).
Gas 2: \(n_2 = 3\), \(\gamma_2 = 4/3 \implies C_{v2} = \frac{R}{4/3 - 1} = \frac{R}{1/3} = 3R\).
Calculate mixture \(C_{v,mix}\):
\[ C_{v,mix} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2(1.5R) + 3(3R)}{2 + 3} \]
\[ C_{v,mix} = \frac{3R + 9R}{5} = \frac{12R}{5} = 2.4R \]
Now, calculate \(\gamma_{mix}\) using the relation \(C_p = C_v + R\):
\[ C_{p,mix} = C_{v,mix} + R = 2.4R + R = 3.4R \]
\[ \gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{3.4R}{2.4R} = \frac{34}{24} = \frac{17}{12} \]
\[ \gamma_{mix} \approx 1.4166... \approx 1.42 \]
This matches option (B).
Step 4: Final Answer:
The ratio of specific heats for the gas mixture is calculated by determining the average molar heat capacity at constant volume, leading to \(\gamma \approx 1.42\).