I have a balloon that contains 22 grams of Carbon dioxide. The barometric pressure is 1019 millibars which I converted to 1.006 Atm and the temperature is 299 "K". What is the volume of balloon? I used the ideal gas law to find out the volume in liters of the balloon.
Is my calculation correct?
1.006 Atm times unknown Volume = 0.5 Moles of CO2 times 0.0821 times 299
1.006 Atm times Volume = 12.27
I then took 12.27/1.006 and came up with 12.2 L of the balloon. Is this correct? If not where did I go wrong?
Thank you
To determine the correct volume of the balloon using the ideal gas law, we need to utilize the equation PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R is the ideal gas constant, and T stands for temperature in Kelvin.
First, let's convert the given barometric pressure of 1019 millibars to atmospheres. Since 1 atmosphere is equivalent to 1013.25 millibars, we can calculate:
1019 millibars * (1 atm / 1013.25 millibars) ≈ 1.006 atm
The barometric pressure conversion you provided is correct.
Now, let's break down the ideal gas law equation, substituting the known values:
(1.006 atm)(V) = (0.5 moles CO2)(0.0821 L atm / (mol K))(299 K)
Now, let's solve for V:
V = (0.5 moles CO2)(0.0821 L atm / (mol K))(299 K) / (1.006 atm)
= 12.27 L / 1.006
≈ 12.2 L
After performing the calculations, it seems that your calculation is indeed correct. The volume of the balloon is approximately 12.2 liters.
Therefore, your initial calculation is correct, and the volume of the balloon is 12.2 L.