Author Topic: gas laws..  (Read 6348 times)

Offline astarmathsandphysics

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Re: gas laws..
« Reply #45 on: October 13, 2009, 09:06:51 am »
For 10 min each, at 27 degrees celsius,from 2 identical holes,nitrogen and an unknown gas are leaked into a common vessel of 3 litre capacity. The resulting pressure is 4.18 bar and mixture contains 0.4 mole of nitrogen. WHat is the  molar mass of the unknown gas?
pv=nRT
4.18*10^5*0.003=n*8.31*300 so n=0.503 moles so 0.103 moles unknown gas. The kinetic energies are the same(=3/2kT=1/2mv^2) so v=k/sqrt(m) the unknown gas diffuses at 1/4 the rate of nitogen so its mass must be 16 times greater 16*15=240g/mol
« Last Edit: October 13, 2009, 10:01:43 am by astarmathsandphysics »

nid404

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Re: gas laws..
« Reply #46 on: October 13, 2009, 09:22:06 am »
I didn't get it....do you mind explaining it please....

like why did u take V=0.003?? and the kinetic energy part of it

Offline astarmathsandphysics

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Re: gas laws..
« Reply #47 on: October 13, 2009, 09:57:02 am »
3 litres =3000/1000000 m3
The kinetic energy of a gas molecule is 3/2kT so all the molecules have the same energy. The kinetic energy is given by 1/2mv^2 s0o mv^2 for nitrogen =mv^2 for the unknown gas so m is inversly proprtional to v^2. I will make a correction to my earlier post.

nid404

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Re: gas laws..
« Reply #48 on: October 13, 2009, 04:38:31 pm »
thanks a lot astar :)

nid404

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Re: gas laws..
« Reply #49 on: October 20, 2009, 07:22:40 am »
It's me again ::)

A stopclock, connecting two bulbs of volumes 5l and 10l containing an ideal gas at 9atm and 6atm respectively is opened. What is the final pressure in the two bulbs??

nid404

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Re: gas laws..
« Reply #50 on: October 20, 2009, 07:40:49 am »
And one more

An open flask contains air at 27degrees celsius.Calculate the temperature at which it should be heated so that
a)1/3 of the air measured at 27degrees escapes out
b)1/3 of air measured at final temp escapes out

Offline astarmathsandphysics

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Re: gas laws..
« Reply #51 on: October 20, 2009, 08:10:08 am »
A stopclock, connecting two bulbs of volumes 5l and 10l containing an ideal gas at 9atm and 6atm respectively is opened. What is the final pressure in the two bulbs??

partial pressures question
pressure of gas 1 =5/(5+10)*9=3atm
pressure gas 2 =10/(5+10)*6=4atm
partial =7atm
« Last Edit: October 20, 2009, 08:56:56 am by astarmathsandphysics »

Offline astarmathsandphysics

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Re: gas laws..
« Reply #52 on: October 20, 2009, 08:14:39 am »
An open flask contains air at 27degrees celsius.Calculate the temperature at which it should be heated so that
a)1/3 of the air measured at 27degrees escapes out
b)1/3 of air measured at final temp escapes out

1/3 escapes so volume is 3/2 of original volume so temple is 3/2*(273+27)=450K

I dont understand B. 1/3 of the air is 1/3 of the air does it mean the gas is heated again? If so 3/2*(450)=675K

nid404

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Re: gas laws..
« Reply #53 on: October 20, 2009, 08:23:29 am »
even  i don't know abt the b part

thanks for the other ones tho :)


nid404

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Re: gas laws..
« Reply #54 on: October 20, 2009, 08:32:26 am »
sir can u explain the partial pressure thing

how is 10/15=4

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Re: gas laws..
« Reply #55 on: October 20, 2009, 08:55:28 am »
it is 10/15 * 6atm=4.mole fraction * pressure. ::)

Offline astarmathsandphysics

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Re: gas laws..
« Reply #56 on: October 20, 2009, 08:57:13 am »
I corrected my post

nid404

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Re: gas laws..
« Reply #57 on: October 20, 2009, 08:57:45 am »
it is 10/15 * 6atm=4.mole fraction * pressure. ::)

thanks. I just started with the concept...so i really don't know

Offline astarmathsandphysics

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Re: gas laws..
« Reply #58 on: October 20, 2009, 09:02:34 am »
You calculae the pressure that would be exerted by each gas if it occupied the whole volume then add them

nid404

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Re: gas laws..
« Reply #59 on: October 20, 2009, 09:03:02 am »
You calculae the pressure that would be exerted by each gas if it occupied the whole volume then add them

thanks sir