At approximately, 550 K temperature will 40g of argon gas at 2 atm occupy a volume of 22.4L.
The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases.
PV=nRT
P=2 atm
V=22.4L
n= 40g/40g/mol =1 mol
R = 0.0821 L. atm/mol-K
T=?
Putting the values in the equation:
PV=nRT
2 atm X 22.4L = 1 mol X 0.0821 L. atm/mol-K X T
T= 550 K
Hence, at approximately 550 K temperature will 40g of argon gas at 2 atm occupy a volume of 22.4L.
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