Answer:volume occupied by 40 grams of argon gas (Ar) at standard conditions = 22.443L
Step-by-step explanation:
Using the ideal gas law that
PV =nRT
Where p =pressure,
V = volume,
n = number of moles
R = gas constant, and
T= temperature in Kelvin.
Since Argon is at STP( Standard conditions), This means that temperature is 273.15 K and pressure is 1 atm
Also the gas constant= 0.08206 L atm/K mol . we are using this value because its units match the units of our values to aid easy calculation
So,
Number of moles = Mass/ Molar mass
=40g/39.948 g/mol
Number of moles of Argon=1.0013moles.
Now Bringing
pV= nRT
V= nRT/p
=(1.0013mol x 0.08206 L atm/K mol x 273.15 K ) /1atm
Volume =22.443L