A 10.0 mL container of helium is sealed at 22.0°C and 1.00 atm pressure. What pressure would be exerted by the helium if the container were heated to 220°C, assuming the volume remains constant?

• A. 3.00 atm
• B. 2.00 atm
• C. 0.50 atm
• D. 1.50 atm

Answer: (B)

Verified step by step guidance

1. Identify the initial conditions: The initial temperature (T1) is 22.0°C, which needs to be converted to Kelvin by adding 273.15, resulting in T1 = 295.15 K. The initial pressure (P1) is 1.00 atm.
2. Identify the final conditions: The final temperature (T2) is 220°C, which also needs to be converted to Kelvin by adding 273.15, resulting in T2 = 493.15 K. The volume remains constant.
3. Use the formula for Gay-Lussac's Law, which relates pressure and temperature at constant volume: \( \frac{P1}{T1} = \frac{P2}{T2} \). This equation shows that the ratio of pressure to temperature remains constant.
4. Rearrange the equation to solve for the final pressure (P2): \( P2 = P1 \times \frac{T2}{T1} \). Substitute the known values into the equation.
5. Calculate the final pressure using the rearranged equation. This will give you the pressure exerted by the helium when the container is heated to 220°C.