Calculating Kinetic Energy of a Gas at Different Temperatures

[brake4country]

Homework Statement

A stationary gas has a kinetic energy of 500 J at 30 C. What is its kinetic energy at 60 C?

Homework Equations

KE = 3/2 RT

The Attempt at a Solution

I first converted T to Kelvin = 333K. I substituted all the values into find KE but my answer is around 40K, which is not even close to the right answer. KE = 3/2RT = 3/2 * 0.08*333 = 40 K. What am I doing wrong?

 

[Shinaolord]

Isn't $R \approx 8.31$ , not 0.08?

Also isn't the formula you need
$\Delta K_{energy} = \frac{3}{2} R\ \Delta T$, where $\Delta$ means the difference between final and initial.

 

[brake4country]

Right. For these problems, I meant to use R = 8.314 J/moleK. Even if I used that one, I end up with a number huge = 4146 J. The correct answer is 550J

 

[Shinaolord]

Read my edit. You need the changes, not the absolute in kelvin.

 

[brake4country]

I never seen kinetic energy in that way before but yes, that makes sense. I set up the equation as:
500-x = 3/2 (8.31)(-30)
However my final KE = 873.

 

[Borek]

I substituted all the values into find KE but my answer is around 40K

No idea what you are doing here. How come the answer to a question asking for energy is in K?

 

[epenguin]

For some reason we might discuss elsewhere students who (I presume) grasped simple proportion in early years at school can't do it any more later when it is called chemistry.

It is essential to know and understand whence comes the formula KE = (3/2) RT . If you had been asked to calculate KE from scratch you would need to be given R (or to be able to calculate it from something else you know, like a mole occupies 22.4 l at STP). You can calculate R also from the information given in the exercise.

But here you don't need to even consider R. R is independent of temperature. (3/2) is also the same at all temperatures(!). KE is proportional to absolute temperature. You have a ratio of temperatures, what is the ratio of KEs?

(Added - and this cancelling of things that are constant is already there hidden away in the problem because they haven't told you how much of the gas there is - if it doesn't change you don't need to know that either.)

(A separate point is that the temperature scale, practical origins understandable, is in theoretically rather irrational units, degrees, that depend on theoretically complicated and I believe incompletely understood or calculable properties of particulat substance, water . As Feynman pointed out, we might have defined temperature in joules/mole and in that case R would have been 1).

 

[brake4country]

No idea what you are doing here. How come the answer to a question asking for energy is in K?

I understand that this question is asking for KE at 60 C. I also understand that R is a constant. I also understand that KE is proportional to temperature. Yes, I understand these realtionships. The answer should be in J not K that was my mistake.

However, my question is why can't we use KE = 3/2RT to calculate kinetic energy of a gas at 60 C? There has to be a formula to use if the answer is 550J. I mean if we were to look at KE = 3/2RT we would know that if KE goes up then T goes up but the question is by how much?

 

[Borek]

my question is why can't we use KE = 3/2RT to calculate kinetic energy of a gas at 60 C?

Because the formula is wrong. It holds for exactly 1 mole of gas, and you don't know how much gas there is. Sure, you can calculate number of moles of gas remembering that the correct formula is KE = \frac 3 2 nRT (where n is number of moles), and then plug it back with anew temperature into the same formula, but as epenguin have explained - that's completely unnecessary.

 

[brake4country]

Oh I see the subtlety in that equation. This makes much more sense. When I take KE = 3/2nRT I get n = 0.13. Then substituting into the equation again to find KE under 60 C I get approx. 540 J which is close to 550 J. Thank you for explaining that! My book did not specify that KE = 3/2RT for one mole. That's where my confusion was.