Answer for question in subtitle

[Aimée]

Homework Help
"A helium balloon has a volume of 5.0 L at a pressure of 101.3 kPa. The balloon is released and reaches an altitude of 6.5 km at a pressure of 380 mm Hg. If the gas temperature remains the same, what is the new volume of the balloon? Assume the pressure inside the balloon and outside the balloon are equal."

First of all this has to do with Boyle's Law. ... I suppose I'll copy the class notes for you:
Pressure (P) - Force acting on a given area
P = F/A
Pressure is measured in
atm (atmosphere)
mm Hg (milimeters of Mercury)
torr
kPa (kiloPascals... or something like that)

Standard Pressure is the at sea level (think of this with the fact of 14.7 units of pressure pressing down on every square inch.)
1 atm
760 mm Hg
760 torr
101 kPa

Boyle's Law - volume varies inversely with pressure, provided that the temperature remains constant.
Formula: V1P1 = V2P2

Example: A gas has a volume of 250 mL when the pressure is 790 torr. What would be the volume when the pressure is 740 torr?
V1 = 250 mL
P1 = 790 torr
V2 = ?
P2 = 740 torr
--In the following paraenthasis means multiply, if you didn't know that before hand--

(250mL)(790torr) = (V2)(740torr)

Now you have to combine both sides, so to get V2 by itself you must divide 740 torr from both sides:

V2 = (250mL)(790torr)
. . . . . . . . 740 torr

250 X 790 = 197500
197500 divided by 740 = 266.8918919

So the answer is V2 = 266.9 (Assuming everyone knows what rounding is.)

Note: However there will be problems with different units (P1 and P2; V1 and V2).. For these you'll have to convert. For example:
P1 = 740 torr
P2 = 95 kPa

If the problem doesn't say which to convert to, ten you can convert either one; for this we'll do kPa to torr.

95 kPa | 760 torr | (multiply the top numbers, divide by the bottom) = 715.851485 torr
. . . . . .| 101 kPa |

*To figure out which numbers to use, use the:
1 atm
760 mm Hg
760 torr
101 kPa
And same units go diagonal from one another so they cancel eachother out, like in the example.

And now for the problem

"A helium balloon has a volume of 5.0 L at a pressure of 101.3 kPa. The balloon is released and reaches an altitude of 6.5 km at a pressure of 380 mm Hg. If the gas temperature remains the same, what is the new volume of the balloon? Assume the pressure inside the balloon and outside the balloon are equal."

V1 = 5.0 L
P1 = 101.3 kPa
V2 = ?
P2 = 380 mm Hg

Caution!: Different units with the Pressures; must convert. We'll do kPa to mm Hg.

101.3 kPa | 760 mm Hg | = 762.3 mm Hg
. . . . . . . .| 101 kPa |

V2 = (5.0L)(762.3mm Hg)
. . . . . . . . 380 mm Hg

...Multiply top; divide bottom...

Answer is:

V2 = 10.03026316 mm Hg

[Ryhana]

This is one of those things that is more on the chem side...but I guess you could be taught this in physics (Don't know what they teach in physics, won't know until I take it....which I never will so )

And Sorry if I dispute alot of what you said Aimee....its just I'm still hyper, and bored...and Its been a while since I've had a chem challenge...even if it is gas laws...

Also if there are any errors in math...or even reason that you suspect tell me and I'll check my notes, online resources, etc. til I'm sure whats right and whats wrong...however any math errors notticed probably will be easy to check as all this math is essentially easy math.

The problem itself (Restating just for the fun of it):

"A helium balloon has a volume of 5.0 L at a pressure of 101.3 kPa. The balloon is released and reaches an altitude of 6.5 km at a pressure of 380 mm Hg. If the gas temperature remains the same, what is the new volume of the balloon? Assume the pressure inside the balloon and outside the balloon are equal."

Now STP (Standard Temperature Pressure) for all four is as follows:

101.3 kPa (Kilopascals)
1 atm (atmosphere)
760 mmHg (millimeters of Mercury)
760 Torr

Interesting side note here, Torr is normally not used outside of measuring the pressure of storms or Hurricanes...its based off of the mmHg used by barometers...why they couldn't have just stuck with mmHg don't ask me...

Now quickly looking at the problem say that yes you could basically figure it out with Boyles law...and that is what Aimee used. So just to check her work we'll use it too...

First converting over...well this is actually a lot easier than it seems. We could take 380 mmHg and convert it to kPa, but that would take to much work...(for those wondering it would be 50.65 kPa or with Significant Figures 51 kPa)

so to convert over 101.3 kPa to mmHg all we have to do is look at the STPs and we see the conversion already sitting right there for us

101.3 kPa=760 mmHg. Easy!

now Boyles law is in fact (V₁)(P₁)=(V₂)(P₂)

Now just a real quick mention: to quickly check if your doing this right you can look at your numbers at the end one thing about boyles law is as Pressure Decreases, Volume increases. As Pressure increases, Volume Decreases. If you don't see this happening at the end...well then you KNOW something went wrong...

those variables being V=Volume and P=Pressure
now to plug it all in:

(5.0L)(760mmHg)=(V₂)(380mmHg)

now for getting the answer:

(5.0L)(760mmHg)=(V₂)
(380mmHg)

Now for some simple explaining...since the above is a division...the two mmHg will "Cancel" each other out in a since...now this doesn't mean you ignore them...it just means ignore the mmHg part of them...in effect you should now have this:

(5.0L)(760)
(380)

now to do the math and get the answer:

(5.0L)(760)
(380)

(3800)
(380)

V₂=10L

So after all that it simply just doubled...wow....now I'm going to take this a little bit further...why? BECAUSE I CAN!!!!
Seriously I'm doin this just because I like messing with the numbers and what not....and because I'm already bored out of my mind so doin this may actually entertain me...

those that leave now this is your fate:


^you..................................^me

For those that stay...theres A prize at the End!!! YAY!!!
Warning: Unnesscary math and explaining ahead.

Continue at your own risk!!

I warned you!!

Now for some new notes class!!
First we're going to need to know Temperature which isn't as hard as it sounds...
First to define some variables! YAY!!
P=Pressure
V=Volume
but we already knew that didn't we...
n=moles
R=(I really forget what this one is called but its values are pretty much set in stone)
T=Temperature

Now I said R's values are pretty much set in stone well here they are depending on what pressure unit your using:
atm-0.0821
kPa-8.314
mmHg-62.3
Torr-I'm not going to write this her because in my opinion Torr is a waste of time (Look at mmHg)

now some of you that Know chemistry or at least know the gas laws are probably saying, "I think I know where shes going with this..."

well if your thinking Ideal gas law...or Pvnrt (Pronounced Pivnert) then your right!!

Ideal gas law

PV=nRT

ok this is all nice and dandy, but how will this help us figure out temperature?
EASY!!

Since we already know all the other variables...and yes we do know all the variables...all we have to do is plug them in and solve for T!

ok we already know the pressure is 760mmHg, and that the volume is 5L...now what about the rest....well R is going to be 62.3. So so far we have this:

(760mmHg)(5L)=(n)(62.3)(T)

but wait what about Moles (n)?? We can't solve for T if we don't know moles!! Well we do...its just you don't know that we know...

Since we are dealing with Standard Pressure, we know that 1 mole is equal to 22.4 L so that means that we have 0.22321428571428571428571428571429 moles....wow thats a bit big so lets use Sig figs to cut that down....0.223 moles. Yay! so we now have the moles!!

(760mmHg)(5L)=(0.223)(62.3)(T)

ok time to solve it:

(760)(5)=(0.223)(62.3)(T)

(3800)=(13.8929)(T)

(3800)=(T)
(13.8929)

T=273.52100713313994918267604315874^oK

wow another big number...Time to cut it down a bit...this time its rounding not Sig Fig:

T=273.52^oK

ok so we now know that the gas in the ballon (Which we will assume to also be the same temp as out side) is at 273.52 degrees kelvin (Which is barely above 0 degrees celsius) so its really cold...well its about to get colder...

now to acturately get the new volume at the new height we figuring in temp...well we need the new temp as well!

Now we don't know Volume so Pvnrt won't help us here in finding temp.

well we have two options: Use another gas law or Use some info from an Airforce Junior ROTC instructer.

well we'll be exploring both options....

I don't remember the name for this gas law...but like most laws in sceince its named after the dude that discovered it...

this gas law being:

(P₁)=(P₂)
(T₁)..(T₂)


now we just plug in every thing:

(760) = (380)
(273.52) (T₂)

(380)
(T₂) = 2.7785902310617139514477917519743

(T₂) =380 / 2.7785902310617139514477917519743

(T₂) = 136.76^oK

So according to the Gas law it says our new temp is 136.76 degrees Kelvin.
now before we continue lets see what the info from the Airforce guy tells us.

Now for every 1000 feet you go up temp drops 2 degrees celsius...since celsius and kelvin conversions is only addition and subtraction we know that the same is true for kelvin...

so now for the conversion stuff which...I used a conversion calc for all this so if someone else wants to post the conversions go ahead!

6.5 Km= 21278.5 (Rounded)

going from that we find out that there is 21 1,000 increments there (DUH just look at the number) so we take 21 times it by two and we get 42. That means that the temp has dropped 42 degrees...now lets take our beginning temp and plug this in:

273.52-42= 231.52

wow so we have a HUGE difference between the two...now see this is why we are exploring both options... ~takes a look up~ Wow...if yal are still with me...then wow...you must like learning...or just seeing numbers...or you just like me! If its because you like me then Yay ;D!! if not then !

Ok time to find the new volume!

Now we introduce a new Gas law to you:

This gas law I can't remember the name to either.

(P₁)(V₁)=(P₂)(V₂)
..(T₁)........(T₂)

ok so now all we do is plug in the numbers.

Using the temp from the Gas law:

(760)(5)=(380)(V₂)
(273.52)..(136.76)

(380)(V₂)
(136.76)..=13.893967093235831809872029250457

(380)(V₂)=13.893967093235831809872029250457 * 136.76

(380)(V₂)=1900.1389396709323583180987202925

(V₂)=1900.1389396709323583180987202925 / 380

(V₂)=5.0003656307129798903107861060329

So according to this there would have been very little expansion of the ballon...so the ballon would have remained the same size...(This is of course ignoring atmospheric pressure)

Now for the info given by the airforce dude:

I'm going to skip the whole restate the formula thing (As you get to know me you'll find I like doing this sometimes)

So straight to pluging in the info:

(760)(5)=(380)(V₂)
(273.52)..(231.52)

(380)(V₂)
(231.52)..=13.892951155308569757238958759871

(380)(V₂)=13.892951155308569757238958759871 * 231.52

(380)(V₂)=3216.4960514770400701959637320854

(V₂)=3216.4960514770400701959637320854 / 380

(V₂)=8.4644632933606317636735887686458

So by the info from the Airforce, the ballon will increase in size yet not to the extent in the original problem...

Which one is right??

YOU DECIDE!!!!! MWAHAHAHAHAHAHAHAHAHAHAHAHA~cough~ ~cough~

well this has been fun with math and gas laws with Hyper Girl Ryhana!!

And for those that stuck with me, you get this prize:

A big hug from yours truely!!!!

~Hugs you~

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