# Question #6f8d8

Mar 25, 2015

You simply rearrange the equation so that you get the pressure you want isolated on one side.

The combined gas law equation looks like this

$\frac{P V}{T} = \text{constant}$

You'll use this equation in the form

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$, where

${P}_{1}$, ${V}_{1}$, ${T}_{1}$ - the pressure, volume, and temperature of a gas at an initial state;
${P}_{2}$, ${V}_{2}$, ${T}_{2}$ - the pressure, volume, and temperature of that gas at a different state.

So, let's assume you have a gas that goes from

${P}_{1} = \text{1.0 atm}$, ${V}_{1} = \text{5.0 L}$, and ${T}_{1} = \text{293 K}$, to

${P}_{2} = \text{?}$, ${V}_{2} = \text{3.0 L}$, and ${T}_{2} = \text{303 K}$

To determine the pressure of the gas under those new conditions, use the combined gas law like this

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2 \implies {P}_{2} = {T}_{2} / {T}_{1} \cdot {V}_{1} / {V}_{2} \cdot {P}_{1}$

In this case,

${P}_{2} = \text{303 K"/"293 K" * "5.0 L"/"3.0 L" * "1.0 atm" = "1.7 atm}$