A certain gas occupies 0.418 litres at `27^@C` and 740 mm Hg.
What is the volume at S.T.P. ?

Explanation

To find the volume of a gas at Standard Temperature and Pressure (STP), we can use the ideal gas law, which can be expressed as:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \(P_1\) = initial pressure
- \(V_1\) = initial volume
- \(T_1\) = initial temperature (in Kelvin)
- \(P_2\) = final pressure (at STP)
- \(V_2\) = final volume (at STP)
- \(T_2\) = final temperature (at STP)

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume \(V_1 = 0.418 \, \text{liters}\)
- Initial temperature \(T_1 = 27^\circ C\)
- Initial pressure \(P_1 = 740 \, \text{mm Hg}\)

2. **Convert the Initial Temperature to Kelvin:**
\[ T_1 = 27 + 273 = 300 \, \text{K} \]

3. **Identify the Standard Conditions:**
- Standard pressure \(P_2 = 760 \, \text{mm Hg}\)
- Standard temperature \(T_2 = 273 \, \text{K}\)

4. **Set Up the Equation Using the Ideal Gas Law:**
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

5. **Substitute the Known Values into the Equation:**
\[ \frac{740 \, \text{mm Hg} \times 0.418 \, \text{L}}{300 \, \text{K}} = \frac{760 \, \text{mm Hg} \times V_2}{273 \, \text{K}} \]

6. **Cross-Multiply to Solve for \(V_2\):**
\[ 740 \times 0.418 \times 273 = 760 \times V_2 \times 300 \]

7. **Calculate the Left Side:**
\[ 740 \times 0.418 \times 273 \approx 740 \times 114.594 = 84799.32 \]

8. **Now Solve for \(V_2\):**
\[ V_2 = \frac{84799.32}{760 \times 300} \]
\[ V_2 = \frac{84799.32}{228000} \approx 0.372 \, \text{liters} \]

9. **Final Answer:**
\[ V_2 \approx 0.372 \, \text{liters} \]