Help on RMS question

equate kinetic energy to thermal energy.

Reply 2

Worzo, why could you just not use

\langle c\rangle = \sqrt{\frac{3P}{\rho}}

or

\langle c\rangle = \sqrt{\frac{3\,RT}{M_{m}}}

where

M_{m}

is the molar mass of nitrogen?

Reply 3

Christophicus

18

Dharma

Worzo, why could you just not use

\langle c\rangle = \sqrt{\frac{3P}{\rho}}

or

\langle c\rangle = \sqrt{\frac{3\,RT}{M_{m}}}

where

M_{m}

is the molar mass of nitrogen?

0.5M(v(mean))^2 = 3RT/2 (per mole of gas, so M is the molar mass)
v(mean) = (3RT/M)^0.5

PV = RT for 1 mole of gas
so V = RT/P

density, rho = M/V = MP/RT (enabling derivation of the first equation)

So Worzo is correct and at least using this method gives you a basic understanding of the situation.

Reply 4

Dharma - the equations you stated follow from my method, but the problem is you're not given the molar mass of nitrogen!

Equating kinetic energy with the translational component of the thermal energy:
(1/2)mv² = (3/2)kT
v² = 3kT/m (1)

Now use ideal gas law for one mole :tongue:

V_mole = N_AkT
And:[rho]=mN_A/V_mole

---> m = [rho]*kT/p

Sub in the [rho] given in the question and STP values to find m, and sub this value of m into equation (1).

Reply 5

Worzo

Dharma - the equations you stated follow from my method, but the problem is you're not given the molar mass of nitrogen!

Point taken. :smile:

Just out of interest, what about the equation

\langle c\rangle = \sqrt{\frac{3\mathrm{R}T}{\rho}}

? Why can that not be used?

Worzo

Equating kinetic energy with the translational component of the thermal energy:
(1/2)mv² = (3/2)kT
v² = 3kT/m (1)

Now use ideal gas law for one mole :tongue:

V_mole = N_AkT
And:[rho]=mN_A/V_mole

---> m = [rho]*kT/p

Sub in the [rho] given in the question and STP values to find m, and sub this value of m into equation (1).

I'm sorry, but I don't really what you've done here or the reasoning behind it (even though it's right). Please could you explain this to me in a little bit more detail?

Reply 6

Dharma

Just out of interest, what about the equation

\langle c\rangle = \sqrt{\frac{3\mathrm{R}T}{\rho}}

? Why can that not be used?

You've copied your own equation from earlier incorrectly there, but I get what you mean. The reason you can't use it is because you're given the density at STP, but it wants the RMS speed at 227 degrees C.

However, it's the equivalent of the equation I derived. The problem is to find the new density at 227 degrees C, and to do that you need to use the ideal gas law.

I'm sorry, but I don't really what you've done here or the reasoning behind it (even though it's right). Please could you explain this to me in a little bit more detail?

OK, I'll try.

Firstly, you know the expression for kinetic energy is (1/2)mv².
Also, you should know that the kinetic energy of a gas molecule is given by (3/2)kT.

So, you can make these equal to each other and rearrange to get an expression for v².

v² = 3kT/m (equation 1)

--------------------------------------------------------------
I hope this is clear. As an aside, I'll show that this is equivalent to your expression. Using pV_mole = RT for one mole of gas (ideal gas law).

Also using the fact that density is (total mass of one mole of gas molecules, each mass m) divided by (volume of one mole), or: [rho] = N_Am/V_mole

So rearrange this equation for V_mole = N_Am/[rho], and substitute this into:
pV_mole = RT
p{N_Am/[rho]} = RT
Using R = N_Ak ---> T = pm/k[rho] (equation 2)

Substitute this into equation (1) ---> v² = 3p/[rho]
---------------------------------------------------------------------

So now we need to use equation (1) to find v². We can use equation 2, rearranged for:
m = k[rho]T/p

Substitute in for the STP values in order to find m.

m = k[rho]_stpT_stp/p_stp

Then you can substitute this value of m into equation (1).

Reply 7

Worzo

You've copied your own equation from earlier incorrectly there, but I get what you mean. The reason you can't use it is because you're given the density at STP, but it wants the RMS speed at 227 degrees C.Point taken.

As for the workings, I understand it now. :smile:

It's really good, but quite long!

Reply 8

It wasn't that long when I did it the first time. There's 3 relevant equations, and all you have to do is substitute. Isn't really a massive physics question.

In simple terms:
1) Equate kinetic to thermal energy (main physics/conceptal part)
2) Ideal gas law (had to come in somewhere!)
3) Definition of density (GCSE)

That's all the physics. Then the maths is:
Sub 3 into 2 and 2 and into 1.

Reply 9