Question

A sealed container holding 0.0255 L of an ideal gas at 0.981 atm and 65 ∘ C is placed into a refrigerator and cooled to 41 ∘ C with no change in volume. Calculate the final pressure of the gas.

Answer 1

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:

$p\propto T$

where

p is the gas pressure

T is the absolute temperature

For a gas undergoing a transformation, the law can be rewritten as:

$\frac{p_1}{T_1}=\frac{p_2}{T_2}$

where in this problem:

$p_1=0.981 atm$ is the initial pressure of the gas

$T_1=65^{\circ}+273=338 K$ is the initial absolute temperature of the gas

$T_2=41^{\circ}+273=314 K$ is the final temperature of the gas

Solving for p2, we find the final pressure of the gas:

$p_2=\frac{p_1 T_2}{T_1}=\frac{(0.981)(314)}{338}=0.911 atm$