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?

Question

4095 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-24T15:28:35

The new pressure is 4.41 atm.

Given

The pressure ( $P_{1}$ $P_1$ ) of a gas at a temperature $T_{1}$ $T_1$
A second temperature $T_{2}$ $T_2$

Find

The second pressure $P_{2}$ $P_2$

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.