[fjfj]

Hi friends my department is Material scince and Engineering. my chemistry teacher ask this hard question could you write a solution? please help

Because the average kinetic energies of molecules ofdifferent gases are equal at a given temperature, and thekinetic energy of a molecule is directly proportional to its mass and the square of its velocity, molecules of gases that have lower molecular weights have higher average velocities than molecules of gases that have higher molecular weights. At 25°C, does every molecule of hydrogen (H2) have a higher velocity than every molecule of oxygen (O2)? Would any molecules of hydrogen have velocities near 0 m/s? Would any molecules of oxygen have velocities greater than the average velocity of all the hydrogen molecules?

[fjfj]

At 25°C, does every molecule of hydrogen (H2) have a higher velocity than every molecule of oxygen (O2)?

No, they dont. Average kinetic energy is equal, but velocities depend on conditions and not every molecule is at the same speed.

Would any molecules of hydrogen have velocities near 0 m/s?

They might be. Average kinetic energy is stable, but molecules themselves may change. Based on kinetic theory, upon molecule hit, the kinetic energy is transferred to other molecule.

Would any molecules of oxygen have velocities greater than the average velocity of all the hydrogen molecules?

Yes, they would. Again, the kinetic theory mentions we can not talk about a single molecule.Approximate velocities and energies calculated from all molecules. Kinetic theory states; In a moment, the velocities of all molecules in the same gas are not equal, some are faster and some are slower.

[Borek]

Your answers don't look bad to me, but I would add some remarks, just in case your teacher wants something else

At 25°C, does every molecule of hydrogen (H2) have a higher velocity than every molecule of oxygen (O2)?

No, they dont.

Note: most hydrogen molecules will be faster than oxygen molecules.

Would any molecules of hydrogen have velocities near 0 m/s?

They might be.

But it is very unlikely.

Would any molecules of oxygen have velocities greater than the average velocity of all the hydrogen molecules?

Yes, they would.

Again, while technically possible, number of such molecules will be extremely low.

It is possible to calculate number of molecules with a given speed range (Maxwell-Boltzmann distribution). It may happen that number of molecules will be so small even in large amounts of gas (many moles) expected number of molecules will be below 1. Sadly, my mathematical intuition fails here, I am not even sure about orders of magnitude. It could be an interesting experience to find exact answers to all these questions (not that difficult exercise, mostly plug and chug).

[Enthalpy]

I suppose the teacher expects a numerically argumented answer.

The ratio of the mean speeds is sqrt(32/2)=4.

Every H2 versus every O2 : their distributions overlap. It suffices that one H₂ has half the mean speed and one O₂ twice the mean speed, which is nothing exceptional. The speed distribution is NOT a Boltzmann, but the probability must be like (10^-2)² or (10^-4)² - in any case 10²³ molecules suffice easily.

Near zero m/s: it fully depends on how near to zero - and incidentally on how many molecules. Zero plus minus one m/s is easy: the probability for one molecule is like (1/1100²)³, achieved by 10²³ molecules. Here 1100 is the mean speed along each direction, and roughly speaking, one wants <1m/s in each direction. Or refine with a Boltzmann on each component of the kinetic energy rather than the speed. Same conclusion. But if zero means plus minus 1mm/s it becomes improbable, or if the matter amount is tiny.

One oxygen faster than the averge hydrogen: less easy. It needs 16 times the mean oxygen kinetic energy, and the distribution gets really small. This demands a computation.

The desired answer should use the distributions properly I guess... If Maxwell-Boltzmann is too complicated, try one Boltzmann energy distribution in each speed direction, it's simpler when applicable.