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.
To find the average translational kinetic energy of an oxygen molecule, you can use the formula K_trans = (3/2)RT, where R is the gas constant.
First, let's correct the calculation for the temperature (T) using the ideal gas equation, PV = nRT.
Rearranging the equation, we have T = PV / (nR).
Plugging in the given values:
P = 7.99 atm
V = 5.10 L
n = 2.10 mol
R = 0.0821 L·atm/(K·mol) (the value of the gas constant)
T = (7.99 atm * 5.10 L) / (2.10 mol * 0.0821 L·atm/(K·mol))
T ≈ 98.2 K (rounded to three significant figures)
Now that we have the correct temperature, we can calculate the average translational kinetic energy (K_trans) using the formula:
K_trans = (3/2) nRT
Plugging in the values:
n = 2.10 mol
R = 8.314 J/(mol·K) (the value of the gas constant in SI units)
T = 98.2 K
K_trans = (3/2) * 2.10 mol * 8.314 J/(mol·K) * 98.2 K
K_trans ≈ 612 J (rounded to three significant figures)
So, the average translational kinetic energy of an oxygen molecule under these conditions is approximately 612 J/mol.