The volume of 0.5 mole of gas at 1 atm pressure and `273^(@)C` temperature is

Options

A. 22.4 L
B. 11.2 L
C. 44.8 L
D. 5.6 L

Explanation

To find the volume of 0.5 moles of gas at 1 atm pressure and 273°C temperature, we can use the Ideal Gas Law equation:

### Step-by-Step Solution:

1. **Write down the Ideal Gas Law equation:**
\[ PV = nRT \]
where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature (in Kelvin)

2. **Convert the temperature from Celsius to Kelvin:**
\[ T(K) = T(°C) + 273.15 \]
For 273°C:
\[ T = 273 + 273.15 = 546.15 \, K \]

3. **Identify the values:**
- \( n = 0.5 \, \text{moles} \)
- \( P = 1 \, \text{atm} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 546.15 \, K \)

4. **Substitute the values into the Ideal Gas Law equation:**
\[ 1 \, \text{atm} \cdot V = 0.5 \, \text{mol} \cdot 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \cdot 546.15 \, K \]

5. **Calculate the right side:**
\[ V = \frac{0.5 \cdot 0.0821 \cdot 546.15}{1} \]

6. **Perform the multiplication:**
\[ V = 0.5 \cdot 0.0821 \cdot 546.15 \approx 22.4 \, \text{L} \]

7. **Final Result:**
The volume of 0.5 moles of gas at 1 atm pressure and 273°C temperature is approximately:
\[ V \approx 22.4 \, \text{liters} \]