# How do you convert "255 mm Hg" to "atm"? What about "620 torr" to "PSI"?

May 18, 2016

THE CONVERSION RATIOS

There is a conversion factor for going from $\text{mm Hg}$ (or $\text{torr}$) to $\text{atm}$:

$\setminus m a t h b f \left(\text{1 atm" = "760 mm Hg" = "760 torr}\right)$

There also exists a conversion factor for $\text{PSI}$. I actually don't know it, so I looked it up: https://www.google.com/?gws_rd=ssl#q=psi+to+torr

Turns out it's $\setminus m a t h b f \left(\text{51.7149 torr = 1 PSI}\right)$.

(Google is a great resource. You should practice with it!)

PERFORMING THE CONVERSIONS

(1)

Let's start from the equality we just wrote out.

$\text{1 atm}$ $=$ $\text{760 mm Hg}$

What happens if we divide by $\text{1 atm}$? We get the actual ratio for the conversion.

$\left(1 \cancel{\text{atm")/(1 cancel"atm}}\right)$ $= 1$ $= \text{760 mm Hg"/"1 atm" = "1 atm"/"760 mm Hg}$

This implies that multiplying by this ratio does NOT change the validity of your answer... only the units. (It preserves your numerical information.)

Thus, we get:

$\textcolor{b l u e}{{P}_{\text{atm") = 255 cancel"mm Hg" xx "1 atm"/(760 cancel"mm Hg}}}$

$=$ $\textcolor{b l u e}{\text{0.336 atm}}$

So now you have the same pressure in different units.

(2)

Like before, we have a ratio that is equal to $1$:

$1 = \text{51.7149 torr"/"1 PSI" = "1 PSI"/"51.7149 torr}$

This ratio is used as follows:

$\textcolor{b l u e}{{P}_{\text{torr") = 620 cancel"torr" xx "1 PSI"/(51.7149 cancel"torr}}}$

$=$ $\textcolor{b l u e}{\text{12.0 PSI}}$

Again, the same pressure as $\text{620 torr}$, but in different units.