Author Topic: gas laws..  (Read 6327 times)

Offline astarmathsandphysics

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Re: gas laws..
« Reply #15 on: October 04, 2009, 12:32:02 pm »
another question

0.23g of a volatile solute displaced 112ml of air at  NTP. Calculate the vapour density and the molecular weight of the substabce

density =0.23g/112cm^3 =2.3*10^-4 kg /112*10-6 m^3 =2.05 KG/m^3
 1 mol of a gas occupies 2240ml at ntp so we have 112/2240=0.5 mols.
0.5 mols weighs 0.23g so 1 mole weighs 0.23/0.05=46g
1 mol contains 6.023*10^23 molecules so 1 molecule weighs 0.046/6.023*10^23 kg

nid404

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Re: gas laws..
« Reply #16 on: October 04, 2009, 12:43:06 pm »
answer is 46.02 and 23.01 ???

Offline astarmathsandphysics

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Re: gas laws..
« Reply #17 on: October 04, 2009, 12:49:02 pm »
46 and 23.

nid404

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Re: gas laws..
« Reply #18 on: October 04, 2009, 12:52:06 pm »
isn't density supposed to be in the form g/cm^3......the mass of 1 mole is 46g...so is the density 46g/cm^3....can u just take it like that??

Offline astarmathsandphysics

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Re: gas laws..
« Reply #19 on: October 04, 2009, 12:57:56 pm »
g/cm3 would be 46/22400

Offline astarmathsandphysics

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Re: gas laws..
« Reply #20 on: October 04, 2009, 12:59:34 pm »
cos at ntp 1 mol of gas occupies 22400ml or 22400cm^3

nid404

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Re: gas laws..
« Reply #21 on: October 04, 2009, 01:01:45 pm »
ok got it...thanks sir

nid404

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Re: gas laws..
« Reply #22 on: October 06, 2009, 07:27:42 am »
another 1 :P

The pressure exerted by a 12g of ideal gas at temp t degrees Celsius in a vessel V Litre is 1atm. When the temp is increased by 10 degrees at the same volume, the pressure increases by 10%. Calculate the temp t and voluume V.(molecular weight of gas=120)

ans
t=-173 degrees
V-0.82 L

nid404

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Re: gas laws..
« Reply #23 on: October 06, 2009, 08:23:16 am »
can some1 please answer.....i need to leave for my class in an hour...pleeeeaaaaaaase

Offline astarmathsandphysics

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Re: gas laws..
« Reply #24 on: October 06, 2009, 08:38:04 am »
The pressure exerted by a 12g of ideal gas at temp t degrees Celsius in a vessel V Litre is 1atm. When the temp is increased by 10 degrees at the same volume, the pressure increases by 10%. Calculate the temp t and voluume V.(molecular weight of gas=120)

ans
t=-173 degrees
V-0.82 L
\frac {p_2}{p_1}=\frac {T_2}{T_1}
1.1=\frac {T_1 +10}{T_1}=1+\frac{10}{T_1}
T_1=110K=-173 degrees celsius
0.1 mol of gas so pV=nRT so V=nRT/p=0.1*8.31*100/100000=8.31*10^-4 m3 or 0.83 ltres

nid404

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Re: gas laws..
« Reply #25 on: October 06, 2009, 08:45:39 am »
thanks a lot astar :)


one more please
A balloon filled with helium rises to a certain height at which it gets fully inflated to a volume of 100000 litres. If at this altittude the pressure is 0.02 atm and temp is 268K, what weight of helium is required to fully inflate the balloon

when i used PV=nRT i don't get the right answer

Ans-36.36g

nid404

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Re: gas laws..
« Reply #26 on: October 06, 2009, 03:22:40 pm »
no one answrin :(

Offline Ghost Of Highbury

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Re: gas laws..
« Reply #27 on: October 06, 2009, 03:36:25 pm »
astar is offline.
divine intervention!

nid404

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Re: gas laws..
« Reply #28 on: October 06, 2009, 03:37:12 pm »
i know dude...some1 else could try

Offline Ghost Of Highbury

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Re: gas laws..
« Reply #29 on: October 06, 2009, 03:38:27 pm »
he is online.~
divine intervention!