Question

A gaseous mixture consists of \(4\ \text{g}\) of oxygen and \(4\ \text{g}\) of helium. The ratio \(\dfrac{C_P}{C_V}\) of the mixture is
\[ \text{(}C_P \text{ and } C_V \text{ are molar specific heats of the mixture at constant pressure and constant volume respectively.)} \]

• A. \(\dfrac{29}{13}\)
• B. \(\dfrac{47}{18}\)
• C. \(\dfrac{47}{29}\)
• D. \(\dfrac{18}{13}\)

Hint:

For an ideal gas mixture, \[ C_P=\sum n_i C_{P,i}, \qquad C_V=\sum n_i C_{V,i}. \] Always calculate the number of moles first and then use weighted heat capacities of the constituent gases.

Answer 1

The Correct Option is C

Step 1: Calculate the number of moles of each gas.
For oxygen \((O_2)\), \[ n_{O_2}=\frac{4}{32} =\frac{1}{8}\ \text{mol} \] For helium \((He)\), \[ n_{He}=\frac{4}{4} =1\ \text{mol} \] Therefore, total number of moles is \[ n=\frac{1}{8}+1 =\frac{9}{8}\ \text{mol} \]

Step 2: Write molar specific heats of the constituent gases.
Helium is a monoatomic gas. Hence, \[ C_{V,He}=\frac{3R}{2}, \qquad C_{P,He}=\frac{5R}{2} \] Oxygen is a diatomic gas. Hence, \[ C_{V,O_2}=\frac{5R}{2}, \qquad C_{P,O_2}=\frac{7R}{2} \]

Step 3: Calculate the total \(C_V\) of the mixture.
The total heat capacity at constant volume is \[ C_V^{(\text{total})} = n_{He}C_{V,He} + n_{O_2}C_{V,O_2} \] \[ = 1\left(\frac{3R}{2}\right) + \frac{1}{8}\left(\frac{5R}{2}\right) \] \[ = \frac{3R}{2} + \frac{5R}{16} \] \[ = \frac{24R+5R}{16} \] \[ = \frac{29R}{16} \]

Step 4: Calculate the total \(C_P\) of the mixture.
Similarly, \[ C_P^{(\text{total})} = n_{He}C_{P,He} + n_{O_2}C_{P,O_2} \] \[ = 1\left(\frac{5R}{2}\right) + \frac{1}{8}\left(\frac{7R}{2}\right) \] \[ = \frac{5R}{2} + \frac{7R}{16} \] \[ = \frac{40R+7R}{16} \] \[ = \frac{47R}{16} \]

Step 5: Determine the ratio \(\dfrac{C_P}{C_V}\).
\[ \frac{C_P}{C_V} = \frac{\frac{47R}{16}} {\frac{29R}{16}} \] Cancelling \(R\) and \(16\), \[ \frac{C_P}{C_V} = \frac{47}{29} \]

Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{C_P}{C_V}=\frac{47}{29}} \]