# A gas occupies 5 L at20 °C and 1 atm. What is its volume at 2 atm and 27 °C?

May 28, 2017

The volume will be 2.6 L.

This looks like a case where we can use the combined gas laws:

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

We can rearrange this formula to get

V_2 = V_1 × p_1/p_2 × T_2/T_1

NTP is 1 atm and 20 °C, so

${p}_{1} = \text{1 atm"; V_1 = 5color(white)(l) "L"; T_1 = (20 + 273.15) K = 293.15 K}$
${p}_{2} = \text{2 atm";V_2 = ?; color(white)(m)T_2 = (27 + 273.15) K = 300.15 K}$

${V}_{2} = 5 \textcolor{w h i t e}{l} \text{L" × (1 color(red)(cancel(color(black)("atm"))))/(2 color(red)(cancel(color(black)("atm")))) × (300.15 color(red)(cancel(color(black)("K"))))/(293.15 color(red)(cancel(color(black)("K")))) = "2.6 L}$