# 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?

Apr 24, 2016

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

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{1} / {T}_{1} = {P}_{2} / {T}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

${P}_{1} = \text{5.00 atm"; T_1 = "(25 + 273.15K)" color(white)(ll)= "298.15 K}$
${P}_{2} = \text{?"; color(white)(mmmll)T_2 = "(-10 + 273.15 K)" = "263.15 K}$

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.