# 568 mL of Chlorine 25 C will occupy what volume at -25 C while the pressure remains constant?

May 3, 2018

$\text{473mL}$

#### Explanation:

Using the combined gas law that states that $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

Since our pressure remains constant, we can eliminate P, leaving us with ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

We need our T to be in Kelvin. To convert Celsius to Kelvin, add 273.

${T}_{1} = 25 + 273 = 298 K$
${T}_{2} = - 25 + 273 = 248 K$

Now let's plug in values.

$\text{568mL"/"298K" = V_2/"248K}$

Isolate ${V}_{2}$ by cross multiplying and dividing.

$\text{568mL" * "248K" =V_2 "298K}$

("568mL" * "248K")/ "298K" = V_2

Our K's cancel, leaving us with $\frac{\text{568mL} \cdot 248}{298} = {V}_{2}$

After calculating and using sig figs, we're given $\text{473mL}$