Calculate the volume occupied by 6.6 kg of argon gas at STP.
To calculate the volume occupied by 6.6 kg of argon gas at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure = 1 atm
V = volume
n = number of moles of gas
R = ideal gas constant = 0.0821 L.atm/mol.K
T = temperature = 273.15 K
First, we need to calculate the number of moles of argon gas in 6.6 kg. The molar mass of argon is approximately 39.95 g/mol.
Number of moles = mass / molar mass
Number of moles = 6600 g / 39.95 g/mol
Number of moles = 165.29 mol
Now, we can substitute the values into the ideal gas law equation:
(1 atm) V = (165.29 mol) (0.0821 L.atm/mol.K) (273.15 K)
V = 3692.38 L
Therefore, 6.6 kg of argon gas at STP would occupy approximately 3692.38 liters of volume.