At what temperature will 1.05 g of argon gas exert a pressure of 1.03 atm in a 5.21 L container?
To find the temperature at which 1.05 g of argon gas exerts a pressure of 1.03 atm in a 5.21 L container, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
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 molar mass of argon to convert grams to moles:
Molar mass of argon (Ar) = 39.95 g/mol
n = mass / molar mass
n = 1.05 g / 39.95 g/mol
Next, we substitute the given values into the ideal gas law equation and solve for T:
PV = nRT
T = PV / (nR)
T = (1.03 atm) * (5.21 L) / [(1.05 g / 39.95 g/mol) * (0.0821 L·atm/mol·K)]
T = 5.37 K
Therefore, at a temperature of 5.37 Kelvin, 1.05 g of argon gas will exert a pressure of 1.03 atm in a 5.21 L container.
Use PV = nRT
n = mols = grams/molar mass
Don't forget T must be in kelvin.