To solve this problem, we need to use the principles of the ideal gas law, specifically the relationship between volume and temperature at constant pressure. This relationship can be expressed as:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Where:
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( T_1 \)[/tex] is the initial temperature in Kelvin
- [tex]\( V_2 \)[/tex] is the final volume
- [tex]\( T_2 \)[/tex] is the final temperature in Kelvin
Given:
- Initial volume ([tex]\( V_1 \)[/tex]) = 3.0 liters
- Initial temperature ([tex]\( T_1 \)[/tex]) = 20°C
First, we need to convert the initial temperature from Celsius to Kelvin, since the ideal gas law requires absolute temperatures:
[tex]\[ T_1 = 20 + 273.15 = 293.15 \text{ K} \][/tex]
The final volume ([tex]\( V_2 \)[/tex]) is given as 4.4 liters. We need to find the final temperature ([tex]\( T_2 \)[/tex]).
Using the formula:
[tex]\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \][/tex]
Plugging in the values:
[tex]\[ \frac{3.0 \, \text{L}}{293.15 \, \text{K}} = \frac{4.4 \, \text{L}}{T_2} \][/tex]
Solving for [tex]\( T_2 \)[/tex]:
[tex]\[ T_2 = \frac{4.4 \times 293.15}{3.0} \][/tex]
[tex]\[ T_2 \approx 429.953 \, \text{K} \][/tex]
Finally, convert the final temperature from Kelvin back to Celsius:
[tex]\[ T_2 = 429.953 \, \text{K} - 273.15 \][/tex]
[tex]\[ T_2 \approx 156.803 \, \text{°C} \][/tex]
Therefore, the new temperature is approximately 157°C. The correct answer is:
C. 157°C
