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

0.911 atm

Step-by-step 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:

`(p_1)/(T_1)=(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=(p_1 T_2)/(T_1)=((0.981)(314))/(338)=0.911 atm`