A2 thermal physics past paper help

[greener1993]

Im don't even know where to start, theses questions, it doesn't seem to link to anything I've learned or any equations. So here we go.

1a) show that about 78,000 J are required to move 28 grams of notrogen molecules (one mole) from the surface of the moon to a point far away from the surface.
I have an image which i can't upload because of the restrictions of size, but it has the moon at the bottom then height above the surface varing with gravitstionsl potential energy. Been as it wants it far from the moon ill take the furthest one away which is Height = 7.2*10^5 and GP of -2.0*10^6

1b)Each molecule has a mass of 4.7*10^-26kg. In one mole there are 6.0*20^23 nitrogen molecules. Show that the speed required for molecules at the surface of the planet to escape is about 2400ms.

2) Use the gas law and pv=1/3Nmc^2 to show that the mean square speed of a molecule of an ideal gas at absolute temperature T is given by:
C^2=3RT/Mm where Mm= the mass of one mole of gas.
I got somewhere with this.
1/3NMc^2=NRT
Nmc^2=3nRT, not very far but at least i reconised it which made me depressed because i knew this question wasnt put in by mistake :P

3)The mean surface Temp of the moon is 290K.
Calulate the root mean square speed of nitrogen molecules at 290k.

4) explain why the moon has lost any nitrogen atmosphere it may have once possessed?

Like i said apart from 2 and a word in 3 i don't reconise it, so help me out :P

 

[Delphi51]

1a) The gravitational energy that mass m has due to the gravity of mass M located distance R away is U = -GMm/R.
See http://en.wikipedia.org/wiki/Gravitational_potential_energy#Gravitational_potential_energy
Notice that this formula gives zero energy when R is infinity, and negative energies when R is smaller. You must figure out U for the radius of the moon (m=28 grams, M = mass of moon) and then subtract that value from zero (which eliminates the minus sign) to get the energy required to move m to infinite distance.

b) Now that you know from (a) the energy needed to make 28 grams escape from the moon, can you find the energy needed to make one molecule escape? Just a proportion to solve.

2) I think you have this solved. Note that Mm is the same as m, so just solve for c². Kind of odd to use the letter c for velocity.

3) Use the formula you found in 2) to find the velocity of nitrogen molecules on the moon.

4) Compare answers for 3) and 1b). Is the nitrogen going fast enough to escape the moon?

 

[greener1993]

1 a + b) i still don't understand it mate, I've never seen that equation before, nor is it in the formula sheet, I don't think they would assume you to know that a A level. There must be another way, however even with that equation i can't swing my head around it.

2) I don't think i have solved it, where can i go from the position i got to? P.S yes Ocr B is the worst exam bored ever made, there resources are rubbish and limited and its common to use stupidly werid symbols and change them every 10 pages or so.

3) :) got that, seems obvious now :) thankyou, got 2.6*10^5, stupidly big so i hope it right

4) yes that higher than 2.6*10^5 and is greater than 2400 and will escape :)

Cheers for the help :)

 

[Delphi51]

1. I can't think of another way.
2. You have Nmc^2=3NRT. Divide both sides by Nm to solve for c².
Most welcome.

 

[greener1993]

2) sorry i believe your confused, Nmc^2=3nRT
N=number of molecules
n=number of moles
It may be my fault i might have put it in that format. So dividing by nm will give 3nrt/NM not what i need, i need to somehow make Mm which is mass per one mole from mole and mass.

 

[Delphi51]

The formula is supposed to be c² = 3RT/M, M= molar mass.
(see http://science.uwaterloo.ca/~cchieh/cact/c120/gaskinetics.html)

This works out: PV = nRT = PV=1/3Nmc^2.
Replace Nm with nM: nRT = 1/3nMc²
c² = 3RT/M

There appears to be an error in the answer given in the question.