Find the molecular kinetic energy of 1 g of helium at S.T.P. Given `R = 8.3 xx 10^(7) erg`.

Answer

To find the molecular kinetic energy of 1 g of helium at standard temperature and pressure (S.T.P.), we can use the formula for the average kinetic energy of a gas:

\[ KE = \frac{3}{2} nRT \]

where:
- \( KE \) is the kinetic energy,
- \( n \) is the number of moles,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin.

### Step 1: Calculate the number of moles of helium
The molar mass of helium (He) is approximately 4 g/mol. Therefore, the number of moles \( n \) in 1 g of helium can be calculated as:

\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{4 \text{ g/mol}} = 0.25 \text{ mol} \]

### Step 2: Use the value of \( R \) and \( T \)
Given:
- \( R = 8.3 \times 10^7 \text{ erg} \)
- At S.T.P., the temperature \( T = 273 \text{ K} \)

### Step 3: Substitute values into the kinetic energy formula
Now we can substitute the values of \( n \), \( R \), and \( T \) into the kinetic energy formula:

\[ KE = \frac{3}{2} nRT = \frac{3}{2} \times 0.25 \text{ mol} \times 8.3 \times 10^7 \text{ erg} \times 273 \text{ K} \]

### Step 4: Calculate the kinetic energy
Now, we perform the calculation step by step:

1. Calculate \( nRT \):
\[ nRT = 0.25 \times 8.3 \times 10^7 \times 273 \]

2. First, calculate \( 8.3 \times 273 \):
\[ 8.3 \times 273 = 2265.9 \]

3. Now multiply by \( 0.25 \):
\[ 0.25 \times 2265.9 = 566.475 \]

4. Finally, multiply by \( \frac{3}{2} \):
\[ KE = \frac{3}{2} \times 566.475 = 849.7125 \text{ erg} \]

### Final Result
Thus, the molecular kinetic energy of 1 g of helium at S.T.P. is approximately:

\[ KE \approx 8.5 \times 10^2 \text{ erg} \quad \text{(or 849.7125 erg)} \]