RJC prelims 2000 physics S paper Q2

eagle SG

For those who are interested... I digged out my cupboard to find one whole set of prelim papers... I didn't managed to do all question though Sad

Anyway, this is one short question with no picture, so I thought I would share it for those who want to challenge themselves

Q2)

A vessel is divided into two parts of equal volume by a partition in which there is a very small hole. Initially, each part contains gas at 300K and a low pressure, p. One part of the vessel is now heated to 600K while the other is maintained at 300K. If a steady state is established when the rate at which molecules pass through the hole from each side is the same, find the resulting pressure difference between the two parts.

darkhour SG

can give clue on what the steady rate implies? Laughing

dbowie SG

For some reason physics question seems easy when reading when doing it it becomes farking hard..

darkhour SG

this is S paper leh...means H3 now leh Laughing

tanjun SG

I will like to inquire on this qn. If the two parts are separated by a partition in which there is a very small hole, air can flow freely to both sides. Hence pressure on both sides as time goes by is the same. So, the pressure difference should be zero. But since this qn comes from S paper, I believe that there is more to it.

Anyway, my explanation is that, let's call the two parts of the vessel Part A and Part B.
If Part B is heated from 300K to 600K,
according to Ideal Gas equation PV = nRT
if T is doubled, P is also doubled since V, volume of part B is the same, n, the no of moles of the gas is the same as no additional gas is being pumped into the vessel. Therefore, pressure in Part B initially will be 2p.

However, like I have explained earlier, the small hole causes air to flow freely between two parts.

darkhour SG

agree, thats y there is the 'steady rate' the prob is u haf to figure out what the steady rate implies. moreover i think the heated part will diffuse its molecules at a faster rate, since its molecules haf more energy.

i wuld like to ask something though...the partition isnt movable rite?and also, if there is pressure difference, wun the molecules move over to the other side faster, from the higher pressure to the lower pressure?unless ur saying one side has higher temperature n the other has higher pressure, which may be the solution but i still dun know how to find exact relations b/w those 2=p

eagle SG

I cannot explain physically why there's a pressure difference Sad

But you can imagine this. This hole is small. Molecules are constantly moving, and they can move from one side to the other side no matter the pressure difference. Hence, it is possible that for a certain pressure on the 600K side, the number of molecules moving through the hole is equal to the number of molecules coming through the hole on the other side.

we know that average frequency of molecules hitting one side is v/2L, where v is the velocity of the molecules and L is the length of the cubical volume (assuming it is a cube on each side; it shouldn't matter much though because if it isn't, it is just another constant. We will not be taking into account this constant)

On average, N/3 particles hit 1 side

Hence, the number of molecules through the hole is A/L^2 x N/3 x v/2L
where A/L^2 will be the probability of a molecule reaching the hole.

At steady rate,

Nv on one side will be the same as Nv on the other side.

I stop here in case there's any mistake... Pls help check... I did this question like 6 yrs ago... already cannot really remember... Reading from my own answer sheet, which may be wrong because the teacher never went through it.

tanjun SG

Let me explain what I think and perceive as steady rate mentioned by eagle.

One thing to note is the small hole. I shall use a straw as my illustration.

For example, if one holds a straw in the air without sucking it or sumthing, there are already air particles coming in and out of the straw. However, the rate of air molecules coming in must be the same as the rate of air molecules coming outso as to maintain the atmospheric pressure of 1 atm.

If now, I place one end of the straw in my mouth and I suck it, covering the hole on the other end. I am taking away air from the straw, and thus the pressure too. There is a pressure difference now and hence the straw is compressed by external forces.

Now, if I move my hand on the left and right such that there is a small hole at the end of the straw, air will quickly go into the small hole to fill up the vaccuum. A sucking sound will be heard. Pressure inside the straw will increase.

What I get from this illustration is that, since there is this small hole in the partition and there is an initial pressure difference of p between the two parts, similarly, air will rush in from Part B to Part A until an equilibrium is reached when pressure difference bet Part A and B is zero. And if I am not wrong, this equilibrium refers to the steady state in eagle qn.
However, the total pressure for Part A and Part B will be higher.

angelfairy SG

temp higher also implies that KE of molecules is larger hence more collisions with the walls of the container per unit time, each collision also results in greater change of momentum so force per unit area on the walls which is pressure increases for the 600k vs the 300k container

eagle SG

this is included in the v/2L above => higher KE means higher frequency (number of collisions)

Molecules that has KE related to 600K, after passing through the hole, will become molecules with KE related to 300K, and vice versa.

Thus I think because there's a constant temperature difference, there must be a pressure difference... What the straw example above was in the case of same temperature on both side.

Energy is lost and energy is gained from and by external sources....

dumbdumb! SG

lucky i nv go JC

tanjun SG

Yup. I have forgotten by making the straw illustration, I have made an assumption that the temperature is constant. I apologise for the mistake. Anyway, I am thinking that with regards to the above mentioned eqn, how cum v on both sides are the same since temp is different? Please kindly enlighten me.

Or is it a case of water tank? Imagine the air molecules as volume of water in the tank. Pardon me for the poor illustration.

For part B, I attach a tap to pump in water to it and for part A, I attach another tap for drainage to it.

If Part B is heated up from 300K to 600K, it is like pumping in more water into Part B. Pressure in Part B will increase by 2 times as shown in Ideal Gas equation assuming that the gas is ideal. While thinking abt this illustration, I will like to ask whether the temperature of Part A of the vessel is strictly at 300K.

2 Scenarios

1. If Part A temp is strictly at 300K, it means that something is there to cool it. Using the water tanks illustration, it is akin to putting a drainage tap onto Part A. Water will be pumping in to Part B, flows to Part A and drains out. Hence, there will be a constant pressure difference between the two parts.

Regards to the experiment, if Part A temp is strictly controlled at 300K, the flow of faster molecules at B to A is higher than the flow of slower molecules at A to B. Thus, like what you have said earlier, heat is lost and gained by external sources. Thus, there will be a constant pressure difference.

2. If Part A temp initially is at 300K, and there is nothing to control it. Using the water tanks illustration, water is pumped into part B. There is a initial pressure difference. Hence, water will gush into part A. Volume of water in A increases rapidly at first but becomes slower to the end.

Regards to the experiment, if Part A temp is not strictly controlled, similarly to the case above, the flow of faster molecules at B to A is higher than the flow of slower molecules at A to B. Think abt this. Faster molecules at B to A and meanwhile there are also slower molecules in A to B. Molecules get heated up at B. Overall, the average K.E of the molecules is higher. Thus, by right, the temp of Part A will also increase since average K.E. in A is higher now due to higher temp environment in B. There will still be a pressure difference. However, it will not be constant anymore. It will be like a y= 1/x curve. The pressure difference will be high at first then, it got slower at the end.

Lastly, pardon for the long reply. Mr. Green And thanks for the wonderful qn that makes all of us think.

eagle SG

I think that the temperature is maintained at a constant 600K and 300K on each side.

with regards to my equations, v is not a constant. I meant Nv is a constant, which is the number of molecules multiplied by the velocity, when at steady rate.

from Nv being constant, you can derive that nv is a constant, where n is the number of moles.

at steady rate

300K: n1v1 (final num of moles and final velocity in 300K part)
600K: n2v2 (final num of moles and final velocity in 600K part)

n1v1=n2v2

Remember from ideal gas equations, v = sqrt(3RT/M)

you will find that

n1 = sqrt(2) x n2

Let initial num of moles = n on each side

Hence, we can then find n1 and n2

P1V = n1R(300) {PV = nRT}
P2V = n2R(600)

V is same on both sides

Hence, we can then get P1 and P2.

Subtracting gives you the pressure difference

Hope it answers everything

tanjun SG

Thanks for the clarificaions. Laughing

unclebutcher SG

i just wanna say:

WTF

eagle SG

You mean What's The Formulas?

All the ideal gas equations formulas were used... Power of S paper...

Anyone gamed for Olympiad questions? But cannot ask me to exprain one

tanjun SG

Maybe we can give some gd and trick qn that is commonly set in exams in the forums to help those revising for O and A levels.

unclebutcher SG

\

nope.....WTF is WA! That's Fierce! Laughing

eagle SG

That I cannot do already... My A levels was so long ago....