We are given air inside a tire. Since this is a system that is at a constant volume we can use Gay-Lussac's law, which states the following:
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]Now we solve for the second temperature, first by multiplying both sides by T2:
[tex]T_2\times\frac{P_1}{T_1}=P_2[/tex]Now we multiply by T1/P1:
[tex]T_2=T_1\times\frac{P_2}{P_1_{}}[/tex]Now we replace the known values:
[tex]T_2=(20^0C)\frac{(1.9atm)}{1.8atm}[/tex]The temperature must be in Kelvin, therefore, we use the following conversion factor:
[tex]\begin{gathered} K=C+273.15 \\ K=20+273.15 \\ K=293.15 \end{gathered}[/tex]Replacing in the temperature we get:
[tex]T_2=(293.15K)\frac{(1.9atm)}{1.8atm}[/tex]Now we solve the operations:
[tex]T_2=309.4K[/tex]We can use the same conversion factor to go back to centigrades:
[tex]\begin{gathered} C=K-273.15 \\ C=309.4-273.15 \\ C=36.25 \end{gathered}[/tex]Therefore, the final temperature is 36.25°C.
