Question

One mole of an ideal gas having specific heat ratio (\(\gamma\)) of 1.6 is mixed with one mole of another ideal gas having specific heat ratio of 1.4. If \(C_V\) and \(C_P\) are the molar specific heat capacities of the gas mixture at constant volume and pressure, respectively, which of the following is/are correct? (\(R\) denotes the universal gas constant.)

• A. \(C_V = 2.08R\)
• B. \(C_P = 2.9R\)
• C. \(C_P = 1.48C_V\)
• D. \(C_P = 1.52C_V\)

Hint:

For mixtures of ideal gases, average specific heats can be found by taking mole-fraction-weighted means of individual gas constants.

Answer 1

The Correct Option is A, C

Step 1: Recall relationships

For an ideal gas: $$C_P - C_V = R$$ $$\gamma = \frac{C_P}{C_V}$$

From these: $$C_V = \frac{R}{\gamma - 1}$$ $$C_P = \frac{\gamma R}{\gamma - 1}$$

Step 2: Calculate for each gas

For Gas 1 ($\gamma_1 = 1.6$): $$C_{V,1} = \frac{R}{1.6 - 1} = \frac{R}{0.6} = \frac{5R}{3}$$ $$C_{P,1} = \frac{1.6R}{0.6} = \frac{8R}{3}$$

For Gas 2 ($\gamma_2 = 1.4$): $$C_{V,2} = \frac{R}{1.4 - 1} = \frac{R}{0.4} = \frac{5R}{2}$$ $$C_{P,2} = \frac{1.4R}{0.4} = \frac{7R}{2}$$

Step 3: Calculate mixture properties

For a mixture of ideal gases, the total heat capacity is the sum (since we have 1 mole of each):

$$C_V = \frac{n_1 C_{V,1} + n_2 C_{V,2}}{n_1 + n_2} = \frac{1 \cdot \frac{5R}{3} + 1 \cdot \frac{5R}{2}}{2}$$

$$= \frac{1}{2}\left(\frac{5R}{3} + \frac{5R}{2}\right) = \frac{1}{2}\left(\frac{10R + 15R}{6}\right) = \frac{25R}{12}$$

$$C_V = 2.083R \approx 2.08R$$

$$C_P = \frac{n_1 C_{P,1} + n_2 C_{P,2}}{n_1 + n_2} = \frac{1 \cdot \frac{8R}{3} + 1 \cdot \frac{7R}{2}}{2}$$

$$= \frac{1}{2}\left(\frac{8R}{3} + \frac{7R}{2}\right) = \frac{1}{2}\left(\frac{16R + 21R}{6}\right) = \frac{37R}{12}$$

$$C_P = 3.083R$$

Step 4: Check the ratios

$$\frac{C_P}{C_V} = \frac{37R/12}{25R/12} = \frac{37}{25} = 1.48$$

So: $C_P = 1.48 C_V$

Answer: (A) and (C) are correct