The kinetic energy of 1 g molecule of a gas, at normal temperature and pressure, is

Options

A. `0.56xx10^(-4)J`
B. `2.4xx10^(-2)J`
C. `1.3xx10^(2)J`
D. `3.4xx10^(3)J`

Explanation

To find the kinetic energy of a 1 g molecule of a gas at normal temperature and pressure (NTP), we can use the formula for the average kinetic energy of a gas molecule:

### Step-by-Step Solution:

1. **Understand the formula for kinetic energy**:
The kinetic energy (KE) of one mole of gas at NTP is given by:
\[ KE = \frac{3}{2} RT \]
where \( R \) is the universal gas constant and \( T \) is the absolute temperature in Kelvin.

2. **Identify the values of R and T**:
- The value of the universal gas constant \( R \) is \( 8.314 \, \text{J/mol·K} \).
- The standard temperature at NTP is \( T = 273 \, \text{K} \).

3. **Substitute the values into the formula**:
\[ KE = \frac{3}{2} \times 8.314 \, \text{J/mol·K} \times 273 \, \text{K} \]

4. **Calculate the kinetic energy**:
- First, calculate \( 8.314 \times 273 \):
\[ 8.314 \times 273 = 2270.622 \, \text{J/mol} \]
- Now, multiply by \( \frac{3}{2} \):
\[ KE = \frac{3}{2} \times 2270.622 = 3406.933 \, \text{J/mol} \]

5. **Convert the energy to the energy per molecule**:
Since we are interested in the kinetic energy of a 1 g molecule, we need to convert the energy from per mole to per molecule.
- One mole of gas contains \( 6.022 \times 10^{23} \) molecules (Avogadro's number).
- Therefore, the kinetic energy per molecule is:
\[ KE_{\text{per molecule}} = \frac{3406.933 \, \text{J/mol}}{6.022 \times 10^{23} \, \text{molecules/mol}} \approx 5.66 \times 10^{-21} \, \text{J} \]

6. **Final answer**:
The kinetic energy of 1 g molecule of a gas at normal temperature and pressure is approximately \( 5.66 \times 10^{-21} \, \text{J} \).