Ideal gas law question
A 4.0 liter container has two gases inside, neon and argon. It is known that at 18 °C, the total pressure of the combined gases is 0.850 atm. If it is known that there are 0.100 moles of neon in the container, how many moles of argon are in the container?
n= RT/PV
n= (0.0821 LxATM/KxMOL x 291K) / (0.850 ATM x 4.0 L
= 7.026 mol
I am stuck from here. I don't know how to isolate the 0.100 mols of Neon from the Argon. Initially I subtracted 0.100 mols from 7.026 mol but that is not correct.
Note that you inverted the formula (again?).
The formula is PV = nRT and if you solve for n, that will be
n = PV/RT. You can work this two ways.
n(total) = (0.850)(4.0)/(0.08206)(291) = 0.142 total moles. You know there is 0.100 mol neon; therefore, there must be 0.142-0.100 = 0.042 mol Ar.
Second way.
PV = nRT. Solve for P
P = nRT/V and calculate P for the 0.100 mol Ne.
P = (0.1 x 0.08206 x 291)/4.0 = 0.597 atm
Total P = 0.850 atm so PAr must be the difference. That will be 0.850-0.597 = 0.253 atm. Now plug this 0.253 atm for PAr into n = PV/RT = (0.253)(4.0)/(0.08206)(291) = 0.042 mol.
et
Sorry, I don't understand what you mean by "et". Could you please provide more context or clarify your statement?
To find the number of moles of argon in the container, we can use the ideal gas law equation. However, in order to isolate the moles of argon, we need to determine the total number of moles of both neon and argon combined.
From the ideal gas law equation: n = (PV) / (RT)
Given:
V = 4.0 L (volume of the container)
P = 0.850 atm (total pressure of the combined gases)
T = 18 °C = 18 + 273.15 = 291.15 K (temperature)
We can substitute these values into the equation:
n = (0.850 atm * 4.0 L) / (0.0821 L*atm/(K*mol) * 291.15 K)
n = 7.026 mol
Now we know that the total number of moles of both neon and argon combined is 7.026 moles.
Next, we subtract the known number of moles of neon from the total to find the number of moles of argon:
Moles of argon = Total moles - Moles of neon
Moles of argon = 7.026 mol - 0.100 mol
Moles of argon = 6.926 mol
Therefore, there are approximately 6.926 moles of argon in the container.