Answer: The new pressure of the gas sample once the gas temperature in the jar 0.9424 atm.
Step-by-step explanation:
Volume of the gas = 250 mL
Pressure of the gas at
![T_1=P_1](https://img.qammunity.org/2020/formulas/chemistry/high-school/urqik86j3pskpompn9oqa6ad6nm50g4p93.png)
Pressure of the gas at
![T_2=P_2](https://img.qammunity.org/2020/formulas/chemistry/high-school/hv0exzxslrprfu9dwgv00x21o7org8eyc6.png)
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
,
![T_1=20.3^oC=293.3 K](https://img.qammunity.org/2020/formulas/chemistry/high-school/dkhb2zkbiasc1cpapee5haot9mvvgv4t13.png)
,
![T_2=-2^oC=271 K](https://img.qammunity.org/2020/formulas/chemistry/high-school/avr30x6nvqvvk8ajy723hxdtoupbzxhipy.png)
![(P_1)/(T_1)=(P_2)/(T_2)](https://img.qammunity.org/2020/formulas/chemistry/high-school/rk4xg6yrmhego89fatld1qeuepq135sjo6.png)
![(P_1)/(T_1)* T_2=P_2=(1.02 atm)/(293.3 K)* 271 K=0.9424 atm](https://img.qammunity.org/2020/formulas/chemistry/high-school/bvuqzipn6smuuf1p6r7cvibhpcdvvvij8i.png)
The new pressure of the gas sample once the gas temperature in the jar 0.9424 atm.