I have another question. I've tried it and i keep getting it wrong.
The average kinetic energy of a 1.85-g sample of argon gas in a 3.00-L bulb is 1.28e-22 J/atom.
What is the pressure of the gas?
What is the root mean square velocity of the argon atoms under these conditions?
The average kinetic energy per molecule (or atom, in this case) is (3/2) k T, where k is the Boltzmann constant.
k = 1.36*10^-25 atm*L/K
That tells you that T = 627 K
You know that the number of moles present is
n = 1.85 g/39.94 g/mole = 0.0463 moles
Use those values of n and T and the ideal gas law to compute P = nRT/V.
The rms velocity can be obtained from the average kinetic energy and the mass of an Ar atom
Average KE = (1/2)*M^Vrms^2
i get another wrong answer
It would help if you show your work, and what you consider to be the "right" answer
P= (0.0463*0.082*627)/3 = 0.79 atm
my teacher says i'm wrong but he isn't tlling me the right answer either
To determine the pressure of the gas, you can use the ideal gas law equation: P = nRT/V, where P is the pressure, n is the number of moles, R is the ideal gas constant, T is the temperature in Kelvin, and V is the volume.
Given:
n = 0.0463 moles
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 627 K
V = 3.00 L
Using these values, you can plug them into the equation and solve for P.
P = (0.0463 mol * 0.0821 L·atm/(mol·K) * 627 K) / 3.00 L
= 0.798 atm
So the pressure of the gas is approximately 0.798 atm.
Now, to calculate the root mean square (rms) velocity of the argon atoms, you can use the average kinetic energy formula you mentioned: Average KE = (1/2) * M * Vrms^2, where M is the molar mass of argon.
Given:
Average KE = 1.28e-22 J/atom
Molar mass of argon = 39.94 g/mol
To convert the average kinetic energy from J/atom to J/mol, you need to multiply it by Avogadro's number (6.022 x 10^23 atoms/mol), since there are Avogadro's number of atoms in a mole.
J/mol = (1.28e-22 J/atom) * (6.022 x 10^23 atoms/mol)
Now you can plug these values into the equation and solve for Vrms:
(1/2) * (39.94 g/mol) * Vrms^2 = (1.28e-22 J/atom) * (6.022 x 10^23 atoms/mol)
Simplifying:
Vrms^2 = [(1.28e-22 J/atom) * (6.022 x 10^23 atoms/mol)] / [(1/2) * (39.94 g/mol)]
Vrms^2 ≈ 4.8543e5 m^2/s^2
Taking the square root of both sides:
Vrms ≈ √(4.8543e5 m^2/s^2)
Vrms ≈ 6.974 m/s
So, the root mean square velocity of the argon atoms under these conditions is approximately 6.974 m/s.
I hope this explanation helps you understand the process of solving this problem. If you have any further questions, feel free to ask!