How Many Balloons Can Be Inflated from a Helium Cylinder?

[relatively-uncertain]

Hi everyone, I'd really appreciate any help with this problem:

A helium cylinder for the inflation of party balloons hold s 25.0L of gas and is filled to a pressure of 16500kPa at 15 degrees celsius. How many balloons can be inflated from a single cylinder at 30 degrees celsius if the volume of one balloon is 6.5L and each needs to be inflated to a pressure of 10^8 kPa?

I know I'm supposed to be using the PV=nRT equation, but I have no idea as to how to approach this! Would really appreciate if anyone could offer me any help!
Thanks so much!

 

[Chestermiller]

From your equation, now many moles of helium are contained in the cylinder? How many moles of helium are contained in each balloon?

 

[relatively-uncertain]

From your equation, now many moles of helium are contained in the cylinder? How many moles of helium are contained in each balloon?

From PV=nRT , n= 172.36 mol

And for the balloon:
6.5 • 10^8= n•8.31•303
and n= 2.58x10^5 mol

Would this be the right steps to take?
Thanks!

 

[Chestermiller]

From PV=nRT , n= 172.36 mol

And for the balloon:
6.5 • 10^8= n•8.31•303
and n= 2.58x10^5 mol

Would this be the right steps to take?
Thanks!

Does it make sense to you that a balloon would have more moles than the gas cylinder?

 

[Charles Link]

Does it make sense to you that a balloon would have more moles than the gas cylinder?

I think the OP has a typo=each balloon should be inflated to $P=108$ kPa. $\\$ That would make sense, because normally a balloon gets inflated to a pressure just slightly greater than atmospheric pressure which is $P=101$ kPa . $\\$ A pressure of P=10^8 kPa would be out of the question. (A google shows they estimate the pressure at the center of the Earth to be about $3.6 \, \cdot$ 10^8 kPa. ).

 

[relatively-uncertain]

Does it make sense to you that a balloon would have more moles than the gas cylinder?

I don't think so :/ But I am unsure of how to use any other figures in the equation ie. what to do with the data given in this question. What should I be doing instead that will give me the right number of mols in each ?
Thanks

 

[relatively-uncertain]

I think the OP has a typo=each balloon should be inflated to $P=108$ kPa. $\\$ That would make sense, because normally a balloon gets inflated to a pressure just slightly greater than atmospheric pressure which is $P=101$ kPa . $\\$ A pressure of P=10^8 kPa would be out of the question. (A google shows they estimate the pressure at the center of the Earth to be about $3.6 \, \cdot$ 10^8 kPa. ).

I have just checked the question again and am sure that it is 10^8kPa. It does seem more reasonable to be 108 though.

 

[Charles Link]

I have just checked the question again and am sure that it is 10^8kPa. It does seem more reasonable to be 108 though.

It is clearly a misprint then. With that correction, the problem will give a very reasonable result. $\\$ The gas cylinder at a pressure of 16500 kPa, meanwhile, is a reasonable number. That's somewhere between 2000 and 2500 lbs. of pressure (per square inch), and is commercially available with those numbers.

 

[Chestermiller]

The first step in getting the answer right is to get the units correct and to do the arithmetic correctly. Let's see the details of your calculation for the number of moles of gas in the cylinder. Your present answer is woefully low.

 

[Charles Link]

The first step in getting the answer right is to get the units correct and to do the arithmetic correctly. Let's see the details of your calculation for the number of moles of gas in the cylinder. Your present answer is woefully low.

@Chestermiller The answer of $n=172$ moles of gas in the cylinder is approximately what I computed with a quick hand calculation. The OP needs to show his work though. Looks like he is using $R=8.31$ L kPa /(mole-K) , so that he doesn't need to convert any units other than the temperature, which he did correctly.

 

[relatively-uncertain]

The first step in getting the answer right is to get the units correct and to do the arithmetic correctly. Let's see the details of your calculation for the number of moles of gas in the cylinder. Your present answer is woefully low.

I have calculated it from:

16,500 • 25.0= n•8.31•288
And n = 172.36

Thanks

 

[Chestermiller]

I have calculated it from:

16,500 • 25.0= n•8.31•288
And n = 172.36

Thanks

Looks OK.

 

[relatively-uncertain]

Looks OK.

Using 108 kPa, and the equations above I have got an answer of 588 , but I am not sure it is right

Here is my working:

1. Calculated the mol of helium in cylinder at 30 degrees celsius
PV=nRT
n= (16500•25) / (8.31•303)
=163.83 mol

2. Mol of helium in balloons
n=(108•6.5)/(8.31 • 303)
n=0.2788 mol

3. Divide 163.83/0.2788
= 588

 

[Charles Link]

Using 108 kPa, and the equations above I have got an answer of 588 , but I am not sure it is right

Here is my working:

1. Calculated the mol of helium in cylinder at 30 degrees celsius
PV=nRT
n= (16500•25) / (8.31•303)
=163.83 mol

2. Mol of helium in balloons
n=(108•6.5)/(8.31 • 303)
n=0.2788 mol

3. Divide 163.83/0.2788
= 588

The moles of helium in the cylinder is calculated at 15 degrees Celsius=288 K, and n=172 is correct for that. That number doesn't change. Otherwise, this looks correct.

 

[relatively-uncertain]

Okay - thankyou! So it would be:

127.36/ 0.2788 = 456.8
=457 balloons

I have checked the answer and it is 618. If my working out seems okay, would the answers be incorrect? (Considering that the 10^8kPa is most likely a misprint?)
Thankyou!

 

[Charles Link]

Okay - thankyou! So it would be:

127.36/ 0.2788 = 456.8
=457 balloons

I have checked the answer and it is 618. If my working out seems okay, would the answers be incorrect? (Considering that the 10^8kPa is most likely a misprint?)
Thankyou!

172.36/.2788=618. You typed it in wrong.

 

[relatively-uncertain]

172.36/.2788=618. You typed it in wrong.

Oohhh! Thank you so much ! I really appreciate the help!