Question

The average kinetic energy of 1 mole of a gas at 25 degrees Celsius is:

3.72 x 103 J

3.67 x 101 J

1.65 x 103 J

3.12 x 102 J

Answer 1

@mathstudent55

Answer 2

@Photon336

Answer 3

\(\large KE_{avg} = \dfrac{3}{2}kT\)

Answer 4

You need the temperature in Kelvin and the universal gas constant.

Answer 5

Could you please explain more, I don't get this one.

Answer 6

yeah so kinetic energy and temperature are kind of the same thing same thing.

kinetic energy means energy when the particles are in motion.

\[KE = \frac{ 1 }{ 2 }mv^{2}\]

where m = the mass of the particle and v = the velocity

but you can imagine that some particles are not going to all be moving at the same velocity. so, we take the average kinetic energy of all the particles and we end up getting this:

\[KE_{avg} = \frac{ 3 }{ 2 }kT\]

Answer 7

Okay. Continue......

Answer 8

this is actually the definition of temperature. so technically all you need to find out the kinetic energy which is in Joules, is the temperature which you're given and the boltzmann constant

Answer 9

but first off do you understand what I was saying before?

Answer 10

brb

Answer 11

Back, I do understand.

Answer 12

alright great so \[C^{o}+273 = K \] you need to do is convert 25 degrees C to kelvin

Answer 13

\[1.3807*10^{-23}~\frac{ J }{ K } = k \]

\[KE_{avg} = \frac{ 3 }{ 2 }kT\]

Answer 14

so just plug in k and your value for temperature in Kelvins and you'll get your answer

Answer 15

Sorry I had to leave yesterday.
I just noticed that you need to calculate the average kinetic energy per mole,
not per molecule.
The calculation above is the average kinetic energy per molecule.
Then you need to multiply by Avogadro's number to get it per mole.

Here it is again:

\(\large KE_{avg} = kT\) per molecule, where \(\large k = 1.38066 \times 10^{-23} \dfrac{J}{K} \)

Then you need to multiply the result by Avogadro's number \(\large (6.0221 \times 10^{23} \times \dfrac{1}{mol} )\)
to get the energy per mole.

It is simpler to follow this formula to get the answer per mole:

\(\large KE_{avg} = RT\) per mole, where \(\large R = 8.3145 \dfrac{J}{mol \cdot K} \)

Answer 16

Thanks!

Answer 17

Its fine.