What is the volume occupied by 12.9 g of argon gas at a pressure of 1.30 atm and a temperature of 318 K ?
Use PV = nRT
n = grams Ar/molar mass Ar
R = 0.08206 L*atm/mol*K
To find the volume occupied by the argon gas, we can use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L∙atm/(mol∙K))
T = temperature (in Kelvin)
First, we need to find the number of moles of argon gas. We can use the equation:
n = m/M
Where:
m = mass of the gas (in grams)
M = molar mass of the gas (in grams/mol)
The molar mass of argon (Ar) is approximately 39.95 g/mol.
n = (12.9 g) / (39.95 g/mol) ≈ 0.323 mol
Now we substitute the values into the ideal gas law equation:
(1.30 atm) • V = (0.323 mol) • (0.0821 L∙atm/(mol∙K)) • (318 K)
Simplifying the equation:
1.30V = 8.256 L∙atm/K • mol • K
Dividing both sides by 1.30:
V ≈ 6.35 L
Therefore, the volume occupied by 12.9 g of argon gas at a pressure of 1.30 atm and a temperature of 318 K is approximately 6.35 liters.