Question

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.

Answer 1

983.31K

Explanations:

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.

`\begin{gathered} P\alpha T \\ P=kT \\ (P_1)/(T_1)=(P_2)/(T_2) \end{gathered}`

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:

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

Substitute the given parameters into the formula

`\begin{gathered} T_2=(P_2T_1)/(P_1) \\ T_2=(7.35*198)/(1.48) \\ T_2=(1455.3)/(1.48) \\ T_2=983.31K \\ \\ \end{gathered}`

Hence the final temperature of the gas will be 983.31K