To solve this problem, we will use the relationship given by the combined gas law, which can be simplified for a rigid, sealed container. For such a container, the volume (V) does not change, hence the relationship between the initial and final state of the gas, using the ideal gas law, can be given as:
[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
Given data:
- Initial pressure ([tex]\(P_1\)[/tex]) = 203 kPa
- Initial temperature ([tex]\(T_1\)[/tex]) = 203 K
- Final temperature ([tex]\(T_2\)[/tex]) = 273 K
We need to find the final pressure ([tex]\(P_2\)[/tex]).
Rearrange the equation to solve for [tex]\(P_2\)[/tex]:
[tex]\[ P_2 = P_1 \times \frac{T_2}{T_1} \][/tex]
Substitute the given values:
[tex]\[ P_2 = 203 \, \text{kPa} \times \frac{273 \, \text{K}}{203 \, \text{K}} \][/tex]
Perform the division inside the parenthesis:
[tex]\[ \frac{273 \, \text{K}}{203 \, \text{K}} = 1.345320 \][/tex]
Then multiply:
[tex]\[ P_2 = 203 \, \text{kPa} \times 1.345320 \approx 273 \, \text{kPa} \][/tex]
Thus, the final pressure of the air when the container is heated to 273 K is:
[tex]\[ \boxed{273 \, \text{kPa}} \][/tex]
