# What is the pressure of a gas in "atm" if its pressure is "85 kPa"?

May 19, 2015

I can easily remember that ${10}^{5}$ Pa is in $1$ bar, and I also remember that $1$ bar is equal to slightly less than $1$ atm, so overall I just remember this:

$1$ bar $= 1$ atm $\cdot \left(\frac{100000}{101325}\right)$

and since there are ${10}^{5}$ Pa in $1$ bar, this becomes:

$\cancel{100000 P a} = \text{1 atm} \cdot \left(\frac{\cancel{100000 P a}}{101325 P a}\right)$

$1 = \text{1 atm"/"101325 Pa} \implies$ $\text{101325 Pa" = "1 atm}$

Since there are $101325$ Pa in $1$ atm, there are $101.325$ kPa in $1$ atm.

May 19, 2015

$101325 \text{Pa"=1 "atm}$

$101325 \cancel{\text{Pa}}$ x (1 "kPa")/(1000 cancel"Pa")=101.325 "kPa"

Therefore, $101.325 \text{kPa"=1 "atm}$ .

So if you have a gas at a pressure of $85 \text{kPa}$ and want to convert the pressure to atm, you would do the following:

$85 \cancel{\text{kPa}}$ x (1"atm")/(101.325cancel"kPa")=0.84"atm"