Question

The pressure of oxygen gas inside a container is 5.00 atm at 25°C. If the temperature falls to -10°C, what is the new pressure inside the container?

Answer 1

The new pressure is 4.41 atm.

Explanation:

Given

The pressure (`P_1`) of a gas at a temperature `T_1`
A second temperature `T_2`

Find

The second pressure `P_2`

Strategy

A problem involving two gas pressures and two temperatures must be a Gay-Lussac's Law problem.

Gay-Lussac's Law is

`color(blue)(|bar(ul(color(white)(a/a) P_1/T_1 = P_2/T_2color(white)(a/a)|)))" "`

In your problem,

`P_1 = "5.00 atm"; T_1 = "(25 + 273.15K)" color(white)(ll)= "298.15 K"`
`P_2 = "?"; color(white)(mmmll)T_2 = "(-10 + 273.15 K)" = "263.15 K"`

Insert the numbers into the Gay-Lussac's Law expression.

Solution

`P_1/T_1 = P_2/T_2`

`P_2 = P_1 × T_2/T_1 = "5.00 atm" × (263.15 cancel("K"))/(298.15 cancel("K")) = "4.41 atm"`

The new pressure is 4.41 atm.