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

1 Answer
Jul 6, 2016

Answer:

#V~=7.5*L#

Explanation:

#V=(nRT)/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.