Find the kinetic energy of 2 moles of an ideal gas in calories at `27^@C`

Answer

To find the kinetic energy of 2 moles of an ideal gas at 27°C, we can use the formula for the kinetic energy of an ideal gas:

### Step-by-Step Solution:

1. **Identify the Formula**:
The kinetic energy (KE) of an ideal gas is given by the formula:
\[ KE = \frac{3}{2} nRT \]
where:
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Convert Temperature to Kelvin**:
The temperature in Celsius needs to be converted to Kelvin:
\[ T(K) = T(°C) + 273 \]
For \( 27°C \):
\[ T = 27 + 273 = 300 \, K \]

3. **Determine the Values**:
- Number of moles, \( n = 2 \)
- Universal gas constant, \( R = 2 \, \text{calories/(mole·K)} \)
- Temperature, \( T = 300 \, K \)

4. **Substitute the Values into the Formula**:
Now, substitute the values into the kinetic energy formula:
\[ KE = \frac{3}{2} \times 2 \times 2 \times 300 \]

5. **Calculate the Kinetic Energy**:
- First, calculate \( \frac{3}{2} \times 2 = 3 \)
- Then, multiply by \( R \) and \( T \):
\[ KE = 3 \times 2 \times 300 = 1800 \, \text{calories} \]

### Final Answer:
The kinetic energy of 2 moles of an ideal gas at 27°C is **1800 calories**.

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