Question

How many grams of fluorine gas are in 3.28x10^5 L at a pressure of 5.79 kPa and a temperature of 46° C

Answer 1

$PV=nRT$

$n=PV/RT$

$V=3.28*10^5 L = 3.28 * 10^2 m^3$

$T= 319 K$

$P=5.79*10^3 Pa$

$n=5.79*10^3*3.28*10^2/(319*8.314)=$ $716 mole$

$m=n*M=716mole*38g/mole=27208 g$

The answer is 27208g