Questions: A gas sample has an initial pressure of 547 mm Hg and an initial volume of 0.500 L. What is the pressure (in atm) if we decrease the volume of the sample to 225 mL? Assume the temperature and number of moles remain constant.

Solution Steps

We are given:

• Initial pressure, \( P_1 = 547 \) mm Hg
• Initial volume, \( V_1 = 0.500 \) L
• Final volume, \( V_2 = 225 \) mL

We need to find the final pressure, \( P_2 \), in atm.

Convert the final volume from mL to L: \[ V_2 = 225 \, \text{mL} \times \frac{1 \, \text{L}}{1000 \, \text{mL}} = 0.225 \, \text{L} \]

Boyle's Law states that for a given mass of gas at constant temperature, the product of pressure and volume is constant: \[ P_1 V_1 = P_2 V_2 \]

Rearrange the equation to solve for \( P_2 \): \[ P_2 = \frac{P_1 V_1}{V_2} \]

Substitute the known values: \[ P_2 = \frac{547 \, \text{mm Hg} \times 0.500 \, \text{L}}{0.225 \, \text{L}} \]

\[ P_2 = \frac{273.5 \, \text{mm Hg}}{0.225} \approx 1215.5556 \, \text{mm Hg} \]

Convert the pressure from mm Hg to atm using the conversion factor \( 1 \, \text{atm} = 760 \, \text{mm Hg} \): \[ P_2 = \frac{1215.5556 \, \text{mm Hg}}{760 \, \text{mm Hg/atm}} \approx 1.5994 \, \text{atm} \]

Final Answer