Answer: The new pressure of the gas sample once the gas temperature in the jar 0.9424 atm.
Explanation:
Volume of the gas = 250 mL
Pressure of the gas at [tex]T_1=P_1[/tex]
Pressure of the gas at [tex]T_2=P_2[/tex]
At the constant volume pressure of the gas is directly proportional to the temperature of the gas in Kelvins that is Gay-Lussac's Law
[tex]P_1=1.02 atm[/tex] , [tex]T_1=20.3^oC=293.3 K[/tex]
[tex]P_2=?[/tex] , [tex]T_2=-2^oC=271 K[/tex]
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
[tex]\frac{P_1}{T_1}\times T_2=P_2=\frac{1.02 atm}{293.3 K}\times 271 K=0.9424 atm[/tex]
The new pressure of the gas sample once the gas temperature in the jar 0.9424 atm.
