A sample of gas in a steel container at -75 degrees Celcius and 1.48 atm. At what temperature (in Kelvin) will the sample have a pressure of 7.35 atm? The volume remains constant.

According to the Gay Lussac's law, the pressure of a given mass of gas is directly proportional to the volume provided that the volume is constant.

[tex]\begin{gathered} P\alpha T \\ P=kT \\ \frac{P_1}{T_1}=\frac{P_2}{T_2} \end{gathered}[/tex]

where:

P1 and P2 are the initial and final pressure respectively

T1 and T2 are the initial and final temperature temperature respectively

Given the following parameters:

[tex]\begin{gathered} P_1=1.48atm \\ T_1=-75^0C+273=198K \\ P_2=7.35atm \end{gathered}[/tex]

Substitute the given parameters into the formula

[tex]\begin{gathered} T_2=\frac{P_2T_1}{P_1} \\ T_2=\frac{7.35\times198}{1.48} \\ T_2=\frac{1455.3}{1.48} \\ T_2=983.31K \\ \\ \end{gathered}[/tex]

Hence the final temperature of the gas will be 983.31K