What is the K.E. of one mole of a gas at `237^(@)C` ? Given Boltzamann's constant k `= 1.38 xx 10^(-23) J //k`, Avogadro number `= 6.033 xx 10^(23) `.

Explanation

To find the kinetic energy (K.E.) of one mole of a gas at 237°C, we can use the formula:

\[ \text{K.E.} = \frac{3}{2} n k_B T \]

where:
- \( n \) is the number of moles (which is 1 mole in this case),
- \( k_B \) is Boltzmann's constant,
- \( T \) is the temperature in Kelvin.
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