# Question 12ce4

May 26, 2017

The new pressure is 19.9 psi.

#### 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

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

1. Solve the Gay-Lussac's Law expression for ${p}_{2}$.

2. Insert the numbers into the ${p}_{2}$ expression.

Solution

p_2 = p_1 × T_2/T_1

${p}_{1} = \text{22.5 psi"; T_1 = "(110 + 273.15K)" = "383.15 K}$
p_2 = ?; color(white)(mmm)T_2 = "(65 + 273.15 K)" = "338.15 K"#
${p}_{2} = \text{22.5 psi" × (338.15 cancel("K"))/(383.15 cancel("K")) = "19.9 psi}$