Four moles of hydrogen , two moles of helium and one mole of water vapour form an ideal gas mixture. What is the molar specific heat at constant pressure of mixture?

Options

A. `(16)/(7)R`
B. `(23)/(7)R`
C. `(19)/(7)R`
D. `(26)/(7)R`

Explanation

To find the molar specific heat at constant pressure (\(C_p\)) of the gas mixture consisting of 4 moles of hydrogen, 2 moles of helium, and 1 mole of water vapor, we will follow these steps:

### Step 1: Identify the number of moles and specific heat capacities of each gas
- **Hydrogen (\(H_2\))**: 4 moles, \(C_{p,H_2} = \frac{7}{2} R\) (diatomic gas)
- **Helium (\(He\))**: 2 moles, \(C_{p,He} = \frac{5}{2} R\) (monoatomic gas)
- **Water vapor (\(H_2O\))**: 1 mole, \(C_{p,H_2O} = 4R\) (polyatomic gas)

### Step 2: Use the formula for molar specific heat of a mixture
The formula for the molar specific heat at constant pressure of a mixture is given by:

\[ C_p = \frac{N_1 C_{p1} + N_2 C_{p2} + N_3 C_{p3}}{N_1 + N_2 + N_3} \]

Where:
- \(N_1, N_2, N_3\) are the number of moles of each gas.
- \(C_{p1}, C_{p2}, C_{p3}\) are the specific heats at constant pressure of each gas.

### Step 3: Substitute the values into the formula
Substituting the values we have:

\[ C_p = \frac{(4 \cdot \frac{7}{2} R) + (2 \cdot \frac{5}{2} R) + (1 \cdot 4R)}{4 + 2 + 1} \]

### Step 4: Calculate the numerator and denominator
Calculating the numerator:

- For hydrogen: \(4 \cdot \frac{7}{2} R = 14R\)
- For helium: \(2 \cdot \frac{5}{2} R = 5R\)
- For water vapor: \(1 \cdot 4R = 4R\)

Adding these together:

\[ 14R + 5R + 4R = 23R \]

Calculating the denominator:

\[ 4 + 2 + 1 = 7 \]

### Step 5: Final calculation of \(C_p\)
Now substituting back into the formula:

\[ C_p = \frac{23R}{7} \]

### Conclusion
Thus, the molar specific heat at constant pressure of the mixture is:

\[ C_p = \frac{23R}{7} \]