What is the volume of 9.783 x 1023 atoms of kr at 9.25 atm and 512k?

1 mole of Kr contains Avogadro's number of atoms, therefore, to calculate the number of moles containing 9.783 * 10^23 atoms, we will do cross multiplication as follows:

number of moles = (9.783 * 10^23 * 1) / (6.022 * 10^23) = 1.625 moles

1 atm is equivalent to 101325 Pa, therefore:

9.25 atm = 937256.25 Pa

The ideal gas law states that:

PV = nRT where

P is the pressure = 937256.25 Pa

V is the volume we want to calculate

n is the number of moles = 1.625 moles

R is the gas constant = 8.31441

T is the temperature = 512 k

Substitute with the givens to get V as follows:

937256.25 V = 1.625 * 8.31441 * 512

Volume = 7.38 * 10^-3 liters