The internal energy of one geam of helium at 100 K and one atmospheric pressure is

Options

A. 100 J
B. 1200 J
C. 300 J
D. 500 J

Explanation

To calculate the internal energy of 1 gram of helium at 100 K and 1 atmospheric pressure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Molar Mass of Helium**:
- The molar mass of helium (He) is 4 g/mol.

2. **Convert Mass to Moles**:
- Since we have 1 gram of helium, we need to convert this mass into moles.
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{1 \text{ g}}{4 \text{ g/mol}} = \frac{1}{4} \text{ moles} \]

3. **Use the Formula for Internal Energy**:
- The internal energy (U) for an ideal gas can be calculated using the formula:
\[ U = \frac{3}{2} nRT \]
where:
- \( n \) = number of moles
- \( R \) = universal gas constant (\( R = 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature in Kelvin

4. **Substitute Values into the Formula**:
- Now, substituting the values we have:
\[ U = \frac{3}{2} \left(\frac{1}{4} \text{ moles}\right) (8.314 \, \text{J/(mol K)}) (100 \, \text{K}) \]

5. **Calculate the Internal Energy**:
- First, calculate \( \frac{3}{2} \times \frac{1}{4} = \frac{3}{8} \).
- Now, calculate:
\[ U = \frac{3}{8} \times 8.314 \times 100 \]
\[ U = \frac{3 \times 831.4}{8} = \frac{2494.2}{8} = 311.775 \, \text{J} \]

6. **Final Result**:
- The internal energy of 1 gram of helium at 100 K and 1 atmospheric pressure is approximately 311.775 J.

### Summary:
The internal energy of 1 gram of helium at 100 K and 1 atmospheric pressure is approximately **311.775 Joules**.