How many moles of gas does it take to occupy 120 L at a pressure of 233 kPa and a temperature of 340 K?
To find the number of moles of gas, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in kPa)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K or 8.314 J·mol^-1·K^-1)
T = Temperature (in Kelvin)
First, let's convert the pressure from kPa to atm using the conversion factor: 1 atm = 101.325 kPa.
So, the pressure becomes:
233 kPa * (1 atm / 101.325 kPa) = 2.299 atm
Next, let's convert the temperature from degrees Celsius to Kelvin by adding 273.15.
340 K = 67 °C + 273.15
Now, we have:
P = 2.299 atm
V = 120 L
T = 340 K
Substituting these values into the ideal gas law equation: PV = nRT
(2.299 atm) * (120 L) = n * (0.0821 L·atm/mol·K) * (340 K)
Simplifying, we get:
275.88 = n * 27.914
To solve for n, we divide both sides of the equation by 27.914:
n = 275.88 / 27.914
Calculating this value gives us:
n ≈ 9.894 moles
Therefore, it takes approximately 9.894 moles of gas to occupy 120 L at a pressure of 233 kPa and a temperature of 340 K.