sample of argon gas collected at a pressure of 1.11 atm and a temperature of 28.0 oC is found to occupy a volume of 20.3 liters. How many moles of Ar gas are in the sample?
Use PV = nRT and solve for n = number of moles. Don't forget to convert T to kelvin.
To find the number of moles of argon gas in the sample, we can use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure of the gas (in atm)
V = volume of the gas (in liters)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature of the gas (in Kelvin)
First, let's convert the given temperature from Celsius to Kelvin using the equation:
T(K) = T(°C) + 273.15
So, T = 28.0 °C + 273.15 = 301.15 K
Now, we can plug in the values into the ideal gas law equation:
(1.11 atm) × (20.3 L) = n × (0.0821 L·atm/(mol·K)) × (301.15 K)
Simplifying the equation:
22.533 atm·L = 24.82915 n
Now, solve for n by dividing both sides of the equation by 24.82915:
n = 22.533 atm·L / 24.82915 atm·L/(mol·K)
n ≈ 0.907 mol
Therefore, there are approximately 0.907 moles of argon gas in the sample.