What is the volume occupied by 12.3 g of argon gas at a pressure of 1.17 atm and a temperature of 455K?
Would formula be V=T/P?
388.89 L?
Answered below.
To find the volume occupied by the argon gas, you can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of the gas (1.17 atm)
V = volume of the gas (to be determined)
n = the number of moles of gas
R = the ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature of the gas in Kelvin (455K)
First, let's calculate the number of moles of argon gas using the formula:
n = m/M
Where:
n = number of moles
m = mass of the gas (12.3 g)
M = molar mass of argon gas (39.95 g/mol)
n = 12.3 g / 39.95 g/mol ≈ 0.308 moles
Now, we can rearrange the ideal gas law equation to solve for volume (V):
V = nRT / P
Substituting the known values:
V = (0.308 moles) * (0.0821 L·atm/(mol·K)) * (455K) / (1.17 atm)
V ≈ 12.34 L
Therefore, the volume occupied by 12.3 g of argon gas at a pressure of 1.17 atm and a temperature of 455K is approximately 12.34 L.