A mixture of ideal gasses `N_(2)` and He are taken in the mass ratio 14:1 respectively. Molar heat capacity of the mixture at constant pressure is

Options

A. `(19R)/6`
B. `(6R)/19`
C. `(13R)/6`
D. `(6R)/13`

Explanation

To find the molar heat capacity of a mixture of ideal gases \( N_2 \) and He in a mass ratio of 14:1, we can follow these steps:

### Step 1: Define the mass ratio
Given the mass ratio of \( N_2 \) to He is 14:1, we can express the masses as:
- Mass of \( N_2 = 14m \)
- Mass of He = \( m \)

### Step 2: Calculate the number of moles of each gas
Using the formula for the number of moles:
\[ n = \frac{\text{mass}}{\text{molar mass}} \]

- For \( N_2 \):
\[ n_1 = \frac{14m}{28} = \frac{m}{2} \]

- For He:
\[ n_2 = \frac{m}{4} \]

### Step 3: Determine the molar heat capacities of each gas
The molar heat capacity at constant pressure \( C_p \) for a diatomic gas (like \( N_2 \)):
\[ C_{p1} = \left(1 + \frac{5}{2}\right) R = \frac{7R}{2} \]

For a monatomic gas (like He):
\[ C_{p2} = \left(1 + \frac{3}{2}\right) R = \frac{5R}{2} \]

### Step 4: Use the formula for the molar heat capacity of the mixture
The molar heat capacity \( C_p \) of the mixture can be calculated using:
\[ C_p = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 + n_2} \]

Substituting the values we calculated:
\[ C_p = \frac{\left(\frac{m}{2} \cdot \frac{7R}{2}\right) + \left(\frac{m}{4} \cdot \frac{5R}{2}\right)}{\frac{m}{2} + \frac{m}{4}} \]

### Step 5: Simplify the expression
Calculating the numerator:
\[ = \frac{7mR}{4} + \frac{5mR}{8} = \frac{14mR}{8} + \frac{5mR}{8} = \frac{19mR}{8} \]

Calculating the denominator:
\[ = \frac{m}{2} + \frac{m}{4} = \frac{2m}{4} + \frac{m}{4} = \frac{3m}{4} \]

### Step 6: Final calculation
Now substituting back into the equation for \( C_p \):
\[ C_p = \frac{\frac{19mR}{8}}{\frac{3m}{4}} = \frac{19R}{8} \cdot \frac{4}{3} = \frac{19R}{6} \]

Thus, the molar heat capacity of the mixture at constant pressure is:
\[ \boxed{\frac{19R}{6}} \]