To solve this problem, we will use the Ideal Gas Law, specifically applying the relationship between pressure and temperature when the volume and the amount of gas are constant. This relationship is given by:
[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 in Kelvin
- [tex]\( P_2 \)[/tex] is the final pressure
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin
First, we need to convert the temperatures from Celsius to Kelvin, using the formula:
[tex]\[ T(K) = T(°C) + 273.15 \][/tex]
1. Initial temperature:
[tex]\[ T_1 = -196°C + 273.15 = 77.15 \, \text{K} \][/tex]
2. Final temperature (Standard temperature):
[tex]\[ T_2 = 0°C + 273.15 = 273.15 \, \text{K} \][/tex]
Initial pressure is given as:
[tex]\[ P_1 = 0.062 \, \text{atm} \][/tex]
Now, plug these values into the equation:
[tex]\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \][/tex]
Rearrange to solve for [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 = P_1 \times \frac{T_2}{T_1} \][/tex]
Substitute the known values:
[tex]\[ P_2 = 0.062 \, \text{atm} \times \frac{273.15 \, \text{K}}{77.15 \, \text{K}} \][/tex]
Calculate the ratio:
[tex]\[ \frac{273.15}{77.15} \approx 3.54 \][/tex]
Now, calculate [tex]\( P_2 \)[/tex]:
[tex]\[ P_2 \approx 0.062 \, \text{atm} \times 3.54 \approx 0.219 \, \text{atm} \][/tex]
To match the units of the answer choices, we need to convert this pressure from atmospheres to mmHg using the conversion factor [tex]\(1 \text{atm} = 760 \text{mmHg}\)[/tex]:
[tex]\[ P_2 = 0.219 \, \text{atm} \times 760 \, \text{mmHg/atm} \approx 166.44 \, \text{mmHg} \][/tex]
Rounding to the nearest whole number, we get:
[tex]\[ P_2 \approx 166 \, \text{mmHg} \][/tex]
This value is closest to one of the given answer choices:
[tex]\[ \approx 170 \, \text{mmHg} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{170 \, \text{mmHg}} \][/tex]
