To solve this problem, we need to use Gay-Lussac's Law, which states that the pressure of a gas is directly proportional to its absolute temperature, assuming constant volume. The mathematical representation of Gay-Lussac's Law is:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
where:
- [tex]\( P_1 \)[/tex] is the initial pressure,
- [tex]\( T_1 \)[/tex] is the initial temperature,
- [tex]\( P_2 \)[/tex] is the final pressure,
- [tex]\( T_2 \)[/tex] is the final temperature.
We need to perform the following steps:
1. Convert the temperatures from Celsius to Kelvin:
[tex]\[ T(\text{Kelvin}) = T(\text{Celsius}) + 273.15 \][/tex]
2. Set up the equation using the given values:
- [tex]\( P_1 = 380 \)[/tex] mmHg
- [tex]\( T_1 = 127^\circ\text{C} + 273.15 = 400.15 \)[/tex] K
- [tex]\( T_2 = 27^\circ\text{C} + 273.15 = 300.15 \)[/tex] K
3. Solve for the final pressure [tex]\( P_2 \)[/tex]:
[tex]\[ \frac{P_2}{T_2} = \frac{P_1}{T_1} \][/tex]
Rearranging for [tex]\( P_2 \)[/tex] gives:
[tex]\[ P_2 = P_1 \cdot \frac{T_2}{T_1} \][/tex]
4. Substitute the known values into the equation:
[tex]\[ P_2 = 380 \, \text{mmHg} \cdot \frac{300.15 \, \text{K}}{400.15 \, \text{K}} \][/tex]
5. Compute the final pressure:
[tex]\[ P_2 \approx 285.04 \, \text{mmHg} \][/tex]
Therefore, the pressure of the methane gas when it is cooled to 27°C will be approximately 285 mmHg.
The correct answer is:
B. 285 mmHg
