Question:
At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?
Answer:
The new pressure is: 1.896 atm
Explanation:
At standard temperature and pressure, we have:
[tex]P_1 = 1atm[/tex]
[tex]T_1 = 273.15k[/tex]
[tex]V_1 = 3.5L[/tex]
Outside, we have:
[tex]T_2 = 296k[/tex]
[tex]V_2 = 2.0L[/tex]
Required
Determine the new pressure
Using combined gas law, we have:
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
This gives:
[tex]\frac{1 * 3.5}{273.15} =\frac{P_2*2.0}{296}[/tex]
Solve for [tex]P_2[/tex]
[tex]P_2 = \frac{296 * 1 * 3.5}{273.15*2.0}[/tex]
[tex]P_2 = \frac{1036}{546.30}[/tex]
[tex]P_2 \approx 1.896 atm[/tex]
