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 use the formula for average kinetic energy from the kinetic theory of gases:

\[ KE = \frac{3}{2} k T \]

where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature in Kelvin.

### Step 1: Convert temperatures from Celsius to Kelvin
- For \(N_2\) at \(27^\circ C\):
\[ T_1 = 27 + 273 = 300 \, K \]

- For \(O_2\) at \(227^\circ C\):
\[ T_2 = 227 + 273 = 500 \, K \]

### Step 2: Write the formula for the average kinetic energy for both gases
- Average kinetic energy of \(N_2\):
\[ KE_{N_2} = \frac{3}{2} k T_1 = \frac{3}{2} k (300) \]

- Average kinetic energy of \(O_2\):
\[ KE_{O_2} = \frac{3}{2} k T_2 = \frac{3}{2} k (500) \]

### Step 3: Set up the ratio of the average kinetic energies
\[ \text{Ratio} = \frac{KE_{N_2}}{KE_{O_2}} = \frac{\frac{3}{2} k (300)}{\frac{3}{2} k (500)} \]

### Step 4: Simplify the ratio
- The constants \(\frac{3}{2} k\) cancel out:
\[ \text{Ratio} = \frac{300}{500} = \frac{3}{5} \]

### Conclusion
The ratio of the average kinetic energy of \(N_2\) to \(O_2\) is \(3:5\).

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

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