Calculate kinetic energy of a gram molecule of oxygen at `127^(@)C`. Value of Boltzmann constant `= 1.381 xx 10^(-23) J K^(-1)`. Avogardro's number `= 6.022 xx 10^(23)` per gm-mole.

Explanation

To calculate the kinetic energy of a gram-molecule of oxygen at \(127^\circ C\), we can follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The formula to convert Celsius to Kelvin is:
\[ T(K) = T(°C) + 273 \]
Substituting the given temperature:
\[ T = 127 + 273 = 400 \, K \]

### Step 2: Use the formula for kinetic energy
The kinetic energy (\(KE\)) of a gram-molecule of gas can be calculated using the formula:
\[ KE = \frac{3}{2} N k_B T \]
where:
- \(N\) is Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\)),
- \(k_B\) is Boltzmann's constant (\(1.381 \times 10^{-23} \, \text{J/K}\)),
- \(T\) is the absolute temperature in Kelvin.

### Step 3: Substitute the values into the formula
Now substituting the values into the kinetic energy formula:
\[ KE = \frac{3}{2} \times (6.022 \times 10^{23}) \times (1.381 \times 10^{-23}) \times (400) \]

### Step 4: Calculate the kinetic energy
Calculating the above expression:
1. Calculate \(N \times k_B\):
\[ N \times k_B = (6.022 \times 10^{23}) \times (1.381 \times 10^{-23}) = 8.314 \, J/K \]
2. Now multiply by \(T\):
\[ 8.314 \times 400 = 3325.6 \, J \]
3. Finally, multiply by \(\frac{3}{2}\):
\[ KE = \frac{3}{2} \times 3325.6 = 4988.4 \, J \]

### Final Result
Thus, the kinetic energy of a gram-molecule of oxygen at \(127^\circ C\) is approximately:
\[ KE \approx 4988.4 \, J \]