# How to find the initial pressure?

## Here's my work : ${T}_{1} = 74 + 273.15 = 347.15 k$ ${T}_{2} = 120 + 273.15 = 393.15 k$ ${P}_{2} = 1.79 a t m$ ${P}_{1} {T}_{1} = {P}_{2} {T}_{2}$ ${P}_{1} = \frac{393.15 \cdot 1.79}{347.15} = 2.027 a t m$ Could somebody solve this problem ?

Jan 8, 2018

The initial pressure was 1.58 atm.

#### Explanation:

Since the volume is constant but the pressure and temperature are changing, this is an example of Gay-Lussac's Law:

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

We can rearrange the above formula to get

p_1=p_2 × T_1/T_2

${p}_{1} = \text{?";color(white)(mmmm)T_1 =color(white)(ll) "(74 + 273.15) K"color(white)(l) = "347.15 K}$
${p}_{2} = \text{1.79 atm"; color(white)(l)T_2 = color(white)(l)"(120 + 273.15) K" = "393.15 K}$

${p}_{1} = \text{1.79 atm" × (347.15 color(red)(cancel(color(black)("K"))))/(393.15 color(red)(cancel(color(black)("K")))) = "1.58 atm}$

The initial pressure was 1.58 atm.

(You used the wrong formula for Gay-Lussac's Law.)