A 2.10 mol sample of oxygen gas is confined to a 5.10 L vessel at a pressure of 7.99 atm. Find the average translational kinetic energy of an oxygen molecule under these conditions in (j/mol)
When i was working this out i tried using pv=nrt to find the temperature which i got was 275 K.
I the tried usingg K TotTrans=3/2nRT
3/2(2.10mol)(8.31J/mol)(275K)=7.20x10^3 J/mol. This was not the right answer however, help please.
Where did 3/2 nRT come from?
Ktrans=3/2 kT where k is boltzmann's constant.
So find Temp with pV=nRT, then find 3/2 kT
To find the average translational kinetic energy of an oxygen molecule, we can use the equation:
K_totTrans = (3/2) * N * k * T
where K_totTrans is the average translational kinetic energy, N is the number of moles, k is the Boltzmann constant (8.314 J/mol·K), and T is the temperature in Kelvin.
First, let's find the number of molecules using Avogadro's number (6.02214076 x 10^23 molecules/mol):
N = (2.10 mol) * (6.02214076 x 10^23 molecules/mol)
N ≈ 1.32 x 10^24 molecules
Next, let's find the temperature in Kelvin using the ideal gas law equation:
PV = nRT
Rearranging the equation, we have:
T = (PV) / (nR)
T = (7.99 atm) * (5.10 L) / ((2.10 mol) * (0.0821 L·atm/mol·K))
T ≈ 275 K (as you correctly calculated)
Now, we can substitute the values into the equation to find the average translational kinetic energy:
K_totTrans = (3/2) * (1.32 x 10^24 molecules) * (8.314 J/mol·K) * (275 K)
K_totTrans ≈ 7.17 x 10^26 J
Therefore, the average translational kinetic energy of an oxygen molecule under these conditions is approximately 7.17 x 10^26 J/mol.