# What volume will 1.50*mol of an Ideal Gas occupy at 225*K, at a pressure of 231*mm*Hg?

Feb 2, 2017

$V = \frac{n R T}{P} \cong 90 \cdot L$

#### Explanation:

V=(1.50*molxx0.0821*L*atm*K^-1*mol^-1xx225*K)/((231*mm*Hg)/(760*mm*Hg*atm^-1)

=??L

Do the units cancel out to give an answer in $\text{litres}$? They should!

The key to doing problems such as these is to recognize that $1 \cdot a t m \equiv 760 \cdot m m \cdot H g$. A column of mercury is traditionally used in a laboratory to measure pressure (mind you, these days, they are disappearing due to safety concerns).

$\text{1 atmosphere}$ pressure will support a column of mercury $760 \cdot m m$ high. And thus in the problem we use the quotient, $\frac{231 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1}$ to give an answer in $\frac{1}{a t {m}^{-} 1} = \frac{1}{\frac{1}{a t m}} = a t m$ as required.