Ratio of everage kinetic energy of 14 g of `N_2` at `27^oC` to 24 g of `O_2` at `227^oC` is

Options

A. `1 : 5`
B. `2 : 5`
C. `1 : 4`
D. `3 : 5`

Answer

To find the ratio of the average kinetic energy of 14 g of \( N_2 \) at \( 27^\circ C \) to 24 g of \( O_2 \) at \( 227^\circ C \), we can follow these steps:

### Step 1: Convert the masses of the gases to moles
1. **For \( N_2 \)**:
- Molar mass of \( N_2 \) = 28 g/mol
- Moles of \( N_2 \) = \( \frac{14 \, \text{g}}{28 \, \text{g/mol}} = 0.5 \, \text{mol} \)

2. **For \( O_2 \)**:
- Molar mass of \( O_2 \) = 32 g/mol
- Moles of \( O_2 \) = \( \frac{24 \, \text{g}}{32 \, \text{g/mol}} = 0.75 \, \text{mol} \)

### Step 2: Convert the temperatures to Kelvin
1. **For \( N_2 \)**:
- Temperature in Kelvin = \( 27^\circ C + 273 = 300 \, K \)

2. **For \( O_2 \)**:
- Temperature in Kelvin = \( 227^\circ C + 273 = 500 \, K \)

### Step 3: Use the formula for average kinetic energy
The average kinetic energy (\( KE \)) of one mole of a gas is given by the equation:
\[ KE = \frac{3}{2} RT \]
Where \( R \) is the universal gas constant.

### Step 4: Calculate the average kinetic energy for both gases
1. **For \( N_2 \)**:
- \( KE_{N_2} = \frac{3}{2} R (300) \)

2. **For \( O_2 \)**:
- \( KE_{O_2} = \frac{3}{2} R (500) \)

### Step 5: Find the ratio of the average kinetic energies
\[ \text{Ratio} = \frac{KE_{N_2}}{KE_{O_2}} = \frac{\frac{3}{2} R (300)}{\frac{3}{2} R (500)} \]
The \( \frac{3}{2} R \) cancels out:
\[ \text{Ratio} = \frac{300}{500} = \frac{3}{5} \]

### Final Answer
The ratio of the average kinetic energy of 14 g of \( N_2 \) at \( 27^\circ C \) to 24 g of \( O_2 \) at \( 227^\circ C \) is \( \frac{3}{5} \).