# What is the volume of 9.783*10^23 atoms of Kr at 9.25 atm and 512K?

##### 1 Answer
Jul 6, 2016

$V \cong 7.5 \cdot L$

#### Explanation:

$V = \frac{n R T}{P}$
=(9.783xx10^23*cancel"atoms of Kr")/(6.022xx10^23*cancel"atoms of Kr"*cancel(mol^-1))xx0.0821*L*cancel(atm*K^-1*mol^-1)xx512*cancelKxx(1/(9.25*cancel(atm)))=??L

The expression is dimensionally consistent. I wanted an answer in $L$, and the expression gave me such an answer.