Question

Two moles of helium are mixed with $n$ moles of hydrogen. If $\frac{C_P}{C_V} = \frac{3}{2}$ for the mixture, then the value of n is :

• A. 1
• B. 3
• C. 2
• D. 44622

Answer 1

The Correct Option is C

 To solve this problem, we need to understand the mixture of gases and the concept of specific heat capacities. Specifically, we're looking at a mixture of helium (He) and hydrogen (H₂) gases, and we're given the ratio of specific heat capacities, \(\frac{C_P}{C_V}\), for the mixture.

The ratio \(\frac{C_P}{C_V}\) is known as the adiabatic index \( \gamma \) (gamma). For the mixture, this is given as \(\frac{3}{2}\).

Helium (\( \text{He} \)) is a monoatomic gas with \(\gamma = \frac{5}{3}\).

Hydrogen (\( \text{H}_2 \)) is a diatomic gas with \(\gamma = \frac{7}{5}\).

Let \( n_{He} \) be the number of moles of helium, which is 2 moles.

Let \( n_{H_2} \) be the number of moles of hydrogen, which is \( n \) moles.

The formula for the effective \( \gamma \) (gamma) of a mixture of two gases is given by:

\[\gamma_{\text{mix}} = \frac{n_{He}\cdot C_{V,He} \cdot \gamma_{He} + n_{H_2}\cdot C_{V,H_2} \cdot \gamma_{H_2}}{n_{He}\cdot C_{V,He} + n_{H_2}\cdot C_{V,H_2}}\]

To solve this for \( n \):

• For helium (\( \text{He} \)): \( C_{V,He} = \frac{3R}{2} \) and \( \gamma_{He} = \frac{5}{3} \).
• For hydrogen (\( \text{H}_2 \)): \( C_{V,H_2} = \frac{5R}{2} \) and \( \gamma_{H_2} = \frac{7}{5} \).

Substitute these values into the formula for \( \gamma_{\text{mix}} \):

\[\frac{3}{2} = \frac{2 \cdot \frac{3R}{2} \cdot \frac{5}{3} + n \cdot \frac{5R}{2} \cdot \frac{7}{5}}{2 \cdot \frac{3R}{2} + n \cdot \frac{5R}{2}}\]

Solving the equation:

\[\frac{3}{2} = \frac{5R + 7nR}{3R + 5nR}\]

Multiply both sides by the denominator to clear the fraction:

\[3R + 5nR = \frac{3}{2} \cdot (5R + 7nR)\]

Continue to simplify:

\[6R + 10nR = 15R + 21nR\]

Rearrange terms:

\[6R - 15R = 21nR - 10nR \] \[ -9R = 11nR \]\]