How many moles of gas does it take to occupy 120L at 2.3atm and 340K
I disagree with Dr. Russ's method. Use PV=nRT, where P=2.3atm, V=120L, n= unknown moles, R= .08205, T=340K.
Solve for n to get 9.9 mol gas.
Without significant figures, n=9.893536939.
You will need to assume that it is an ideal gas and that 1 mole of any ideal gas occupies 22.4 litres at STP (273 K, 1 atm).
then use P1V1/T1=P2V2/T2
I would convert 120 L to STP and then compare the volume obtained with 22.4 litres as this then gives you the number of moles.
To determine the number of moles of gas, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's gather the values we have:
Pressure (P) = 2.3 atm
Volume (V) = 120 L
Temperature (T) = 340 K
Next, we need to know the value of the ideal gas constant (R). The value of R depends on the units used, so for this calculation, we will use R = 0.0821 L·atm/(mol·K).
Now, we can rearrange the ideal gas law equation to solve for n:
n = PV / RT
Substituting the given values into the equation:
n = (2.3 atm) * (120 L) / (0.0821 L·atm/(mol·K) * 340 K)
Simplifying the equation:
n = (2.3 atm * 120 L) / (0.0821 L·atm/(mol·K) * 340 K)
n = 276 L·atm / (27.944 L·mol^(-1)·K^(-1))
Finally, calculating the value of n:
n ≈ 9.876 moles
Therefore, it takes approximately 9.876 moles of gas to occupy a volume of 120 L at a pressure of 2.3 atm and a temperature of 340 K.