How do I calculate the average kinetic energy of CH4 and and N2 at 273K and 546K?
https://chem.libretexts.org/Core...This link should really help you.Different gases at the same temperature have the same average kinetic energy as has been said there. So, using the first equation mentioned there you can find the average kinetic energy for both gases at 273K and also at 546K.Hope this helps.
Can you calculate the average kinetic energy of one mole and one molecule of hydrogen gas at 300 Kelvin?
Answer: 6.2129196E-21 JoulesHow To: The Kinetic Molecular Theory formula calculates the average kinetic energy of a gaseous molecule at a specific temperature. The formula for the Kinetic Molecular Theory is:KE = 3/2 • kB • Twhere:kB is the Boltzmann contant (1.3806 x 10-23 m2 kg s-2 K-1)T is the Temperature in KelvinKE is the average kinetic energy of the gaseous molecule.
Calculating the average kinetic energy of molecules?
According to kinetic theory of gases the average kinetic energy of the particles is the same for all ideal gases. It is: K = (3/2)∙k∙T with k Boltzmann constant T thermodynamic temperature. So at 261 K both methane and nitrogen molecules have an average kinetic energy of K = (3/2) ∙ 1.38065×10⁻²³ J∙K⁻¹ ∙ 261 K = 5.405×10⁻²¹ J At 547 K their kinetic energy is: K = (3/2) ∙ 1.38065×10⁻²³ J∙K⁻¹ ∙ 547 K = 1.133×10⁻²⁰ J
Calculating average kinetic energy?
Calculate the average kinetic energies of CH4 and N2 molecules at 273K and 546K. I'm having a whole lot of trouble with this, because I've read my textbook, but it's not very clear about average kinetic energy, so I'm very confused. Any help with this problem as well as explaining KEavg would be greatly appreciated!!!
How can I calculate the average kinetic energy of CH4?
The formula you provided gives the ‘molar kinetic energy”. If you want the average kinetic energy, you need to use a different constant: k=1.38 x 10^-23 KE = 3/2 x (kT) = 3/2 x (1.38 x 10^-23 x 261) = 540.3 x 10^-23 = 5.403 x 10^-21 J KE = 3/2 x (kT) = 3/2 x (1.38 x 10^-23 x 547) = 1132.3 x 10^-23 = 1.13 x 10^-20 Joules Similar examples were covered at these sites: http://answers.yahoo.com/question/index?... http://answers.yahoo.com/question/index?... Hope this helps.
Calculate the most probable speed and average kinetic energy per mole for each of the following.?
For the average energy in J it is an easy calculation: Ek(av)=3/2kT. T here is the thermodynamic temperature in Kelvin scale. For Ar and Xe, the answer is the same: The Boltzmann constant being k=1.38x10EXP(-23)J/Kelvin where 10EXP(N) means the number 10 to the exponent N and T=(112+273)Kelvin then, Ek(av)=1.5x1.38x10EXP(-23)J/Kelvin x 385Kelvin or Ek(av)=796x10EXP(-23)J. For the speed's root mean square we have: 1/2MV(squared)=Ek(av) V(rms)=SQR(2Ek(av)/M) where SQR(N) means the square root of N. V(rms)=SQR(1592x10EXP(-23)/M) M is the mass of one single gas molecule. For Ar, M={40x10EXP(-3)kg/[6,02x10EXP(23)]}so V(rms)=SQR(1592x6.02x1000/40)=489m/s For Xe, M={131x10EXP(-3)kg/[6,02x10EXP(23)]} V(rms)=SQR(1592x6.02x1000/131)=270m/s For C3H8 we have: Ek(av)=[1.5x1.38x10EXP(-23)J/Kelvin]x(... For C3H8 we have the mass of a single molecule: M=(3x12+8x1)x10EXP(-3)kg=44x10EXP(-3)k... V(rms)=SQR(2430x6.02x1000/44)=576m/s It should bee noted that I am here calculating the root mean square or, say, the square root of the average squared speed. I do not know which is the level of this specific problem. If you are at the high school, ok. But if you study college physics, then, the most probable speed should be deduced from the derivative of the boltzmann speed distribution for the molecules of a gas and making that expression equals to zero. In this late case, we have for the Boltzmann distribution of speeds: dN/dv = const x V(squared) x eEXP[-MV(squared)/2kT]; where e is the euler number; e=2.718... Taking the derivative of the above expression and forcing it to be zero we have: V(av)=SQR(2kT/M)=SQR[4/3xEk(av)x10EXP(... For Ar: V(av)=SQR[(796x4/3)x6.02x(1000/40)]=39... For Xe: V(av)=SQR(1063x6.02x1000/131)=220m/s For C3H8: V(av)=SQR[(1215x4/3)x6.02x(1000/40)]=4...
What is the average kinetic energy of each molecule of oxygen at 300K?
The average KE of a particle at temperature T, in K, is given byE = 3/2 kTwhere k is Boltzmann’s constant (1.38x10^-23 J/K).So the average kinetic energy of a particle at 300K would beE = 3/2 x 1.38x10^-23 x 300E = 6.2 x 10^-21 J
How is the formula for the average kinetic energy of a gas molecule derived?
You would have to look up the Kinetic Molecular Theory to see that.Essentially, it is a statistical model where individual molecules are assumed to be essentially spheres that have no internal structure and cannot absorb energy into their interior (all collisions are perfectly elastic). Their Volumes are so tiny that they essentially never collide with each other, only the outer walls. they have intermolecular forces that are so weak that they always travel in straight paths, even if they pass by each other very closely.So, we use a theoretical “ideal gas” model and work with that.Molecules have momentum. when they hit the walls, they exert a force on the wall that can be calculated as the rate of change in momentum. essentially, the change in momentum of any given molecule-wall collision multiplied by the frequency with which the molecule collides with the wall.That tiny force is then multiplied by the number of molecules in the container but we cannot know the individual speed of every molecule so we use instead the root-mean-squared speed (what we call the average) as if all the molecules have the same speed. this way, we can calculate the total rate of momentum change and hence the total force the gas exerts on the wall. then divide by the area and you get the total pressure on the wall.That’s the gist of the derivation but the mathematics is a bit more detailed.You can read my course notes for details. These are pitched at a first-year level and is not the most thorough version of this derivation but it is a start.Gas Laws