What is the volume occupied by 40 grams of argon gas (Ar) at standard conditions?

volume occupied by 40 grams of argon gas (Ar) at standard conditions = 22.443L

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

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