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### Topic: Need guidance with Gas law problem involving a hot air balloon  (Read 4466 times)

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#### zerality

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• Gender:  ##### Need guidance with Gas law problem involving a hot air balloon
« on: December 06, 2006, 12:30:54 AM »
Hi,

I need assistance/ verification of my answer to this problem:

You want to fly a hotair balloon over Mt Everest. The pressure at the top of the mountain is 200 torr. You anticipate an air temp above the mountain of -22 degrees F. and want enough gas volume to provide 500kg of lift. If you can heat the air in your balloon to max. temp of 120 degrees C, what gas volume must your balloon have to achieve lift? (molar mass air = 29)

I said:
500000g = mass air displaced - mass of hot air
= (moles x molar mass) - (moles x molar mass)
= (PV/RT x molar mass) - (PV/RT x molar mass)
= [ (200/760)V/(.08206 X 2.4315 X 10^2)] - [ (200/760)V/ (.08206X 3.9315X10^2)]
500000 = .3824798979V - .2365509021V
V =3.43 X 10^6 L

Is my reasoning and answer correct or do I need to think more???

Thank you

#### Borek ##### Re: Need guidance with Gas law problem involving a hot air balloon
« Reply #1 on: December 06, 2006, 04:35:26 AM »
500000g = mass air displaced - mass of hot air
= (moles x molar mass) - (moles x molar mass)
= (PV/RT x molar mass) - (PV/RT x molar mass)

Is my reasoning and answer correct or do I need to think more???

No idea about numbers, but the approach so far looks OK. What molar mass of air is? (I would call it 'equivalent molar mass' to avoid problems).
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#### zerality

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« Reply #2 on: December 06, 2006, 06:33:29 PM »
500000g = mass air displaced - mass of hot air
= (moles x molar mass) - (moles x molar mass)
= (PV/RT x molar mass) - (PV/RT x molar mass)
= [ (200/760)V/(.08206 X 2.4315 X 10^2)] - [ (200/760)V/ (.08206X 3.9315X10^2)]
500000 = .3824798979V - .2365509021V
V =3.43 X 10^6 L

Was/ am I correct in saying that the P and V  of the mass air displaced is equal to the P and V of hot air?

Thanks

#### billnotgatez

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« Reply #3 on: December 09, 2006, 03:03:49 AM »
78.084% ~ N2
20.947% ~ O2
00.934% ~ Ar
00.033% ~ CO2
00.002% ~ Other

molar mass ~
28.96443

it is interesting that you are allowed to round up to 29
« Last Edit: December 10, 2006, 10:45:44 AM by billnotgatez »

#### Borek ##### Re: Need guidance with Gas law problem involving a hot air balloon
« Reply #4 on: December 09, 2006, 06:34:31 AM »
= (PV/RT x molar mass) - (PV/RT x molar mass)
= [ (200/760)V/(.08206 X 2.4315 X 10^2)] - [ (200/760)V/ (.08206X 3.9315X10^2)]

You have dropped molar mass from the equation, I have quoted only symbolic part of the derivation for a purpose.

Quote
Was/ am I correct in saying that the P and V  of the mass air displaced is equal to the P and V of hot air?

Yes.
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#### billnotgatez

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« Reply #5 on: December 09, 2006, 11:54:06 PM »
zerality -