Answer:
0.911 atm
Explanation:
In this problem, there is no change in volume of the gas, since the container is sealed.
Therefore, we can apply Gay-Lussac's law, which states that:
"For a fixed mass of an ideal gas kept at constant volume, the pressure of the gas is proportional to its absolute temperature"
Mathematically:
[tex]p\propto T[/tex]
where
p is the gas pressure
T is the absolute temperature
For a gas undergoing a transformation, the law can be rewritten as:
[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]
where in this problem:
[tex]p_1=0.981 atm[/tex] is the initial pressure of the gas
[tex]T_1=65^{\circ}+273=338 K[/tex] is the initial absolute temperature of the gas
[tex]T_2=41^{\circ}+273=314 K[/tex] is the final temperature of the gas
Solving for p2, we find the final pressure of the gas:
[tex]p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm[/tex]
