# Ideal Gas Law Helium balloon Problem

1. Dec 13, 2006

### chrispsu

Ideal Gas Law Problem :)

1. The problem statement, all variables and given/known data

A balloon is filled with helium at a pressure of 1.0 x 10^5 Pa. The balloon is at a temperature of 23°C and has a radius of 0.13 m.
(a) How many helium atoms are contained in the balloon?

(b) Suppose we increase the number of helium atoms in the balloon by a factor of 5, keeping the pressure and the temperature fixed. By what factor does the radius of the balloon increase?

2. Relevant equations

PV=nRT
i know i need the 4/3pir^3 for the volume as well

3. The attempt at a solution
for part a)
First off converted Temp to Kelvin.
i used PV=nRT to solve for n.
That answer i multiplied by 6.022x10^23. I thought that would be the correct answer but it is not.
for part b)
I did nx5. PV=nRT again but solved for the volume. Divided by 4/3pi and took the third root of that to find the radius. But again my answer did not come out correct.

Any advice to where I might be going wrong would be awesome! Thanks!

2. Dec 13, 2006

### chrispsu

please anyone who has any idea what im doing wrong i would really appreciate it if you could point it out!!

i keep getting the same number and its an online hwk which is telling me the answer is wrong so im stuck

3. Dec 13, 2006

### Kurdt

Staff Emeritus
Your method seems fine. Perhaps you are comitting minor errors in the calculation. If you post the answers you get people can check them for you and perhaps see where you have gone wrong. One problem might be the fact that you say you've changed the temperature to kelvin. What is the value of the gas constant are you using?

4. Dec 13, 2006

### aeroengphys

The value for R that I use is .08206...if that's the same that you use, you need to convert your pressure into atm's first. Then, solve for volume by doing as you said, V = 4/3 pi r^3. Once you have V, convert degrees celcius into kelvin (once again, only if your value for R is .08206, otherwise, work your units to make sure you're doing it right). Then you'll have P,V,R,and T and you can solve for n. Since n is the number of moles, you just need to multiply that answer by avagadros' number to get the number of atoms.

As for b, look at how V and n are related, and see if you can work it out from there...

5. Dec 13, 2006

### chrispsu

ok i did:

PV=nRT

(1x10^5)(.0092)=n(.08206)(296)
for n i am getting 37.8875
after multiplying it by 6.022x10^23 i get 2.28x10^25 which is incorrect. :(

6. Dec 13, 2006

### Kurdt

Staff Emeritus
I used $$PV = NkT$$ where N is the number of molecules.

$$N=\frac{PV}{kT}$$

I get a slightly different answer to you. Try it this way first and see how you go. k = 1.38x10-23 J/K.

7. Dec 13, 2006

### chrispsu

oh wow that worked!!
thanks alot! :D
ive never seen that formula before lol

8. Dec 13, 2006

### Kurdt

Staff Emeritus
The constant k is simply R multiplied by avagadros number. and instead of dealing with moles of gas deals with the number of molecules in a gas.

9. Dec 13, 2006

### Kurdt

Staff Emeritus
Actually to point out where you went wrong, you have stated R = 0.08206 when the value I have in my text is 8.3145 J/(K mol). if you try your original equation with this value it should work out the same.

10. Dec 13, 2006

### chrispsu

oh ok awesome...i dont know why i used that other value of r, but obviously the other person who posted used it too lol

thanks again!! :D

11. Dec 13, 2006

### chrispsu

if you or anyone else doesnt mind helping me with one more. Ive already gotten it wrong on my homework as your only allowed to submit three tries but I still would like to know how to do it correctly.

The problem is: One mole of a monatomic ideal gas has an initial pressure of 220 kPa, an initial volume of 1.2 x 10-3 m3, and an initial temperature of 350 K. Determine the values of the pressure P after each of the following three sequential processes.
(i) a constant-temperature expansion that quadruples its volume

(ii) a constant-pressure compression to its initial volume

there were 6 parts to this question..i got the other 4 correct, but these two I could not figure out the pressures.

For the second one i thought pressure would remain the same since it was constant-pressure. I dont understand why that is not the case. PV=nRT worked on all the other parts but these, if someone could explain why that would be great!! :D

12. Dec 13, 2006

### Kurdt

Staff Emeritus
For part two the pressure should be as it is after part one has been applied. Again perhaps you have just made a mistake in the calculations.

13. Dec 13, 2006

### chrispsu

Doing PV=nRT i get:

i get P= 606kPa?

14. Dec 13, 2006

### Kurdt

Staff Emeritus
I get that too.

15. Dec 13, 2006

### chrispsu

its not right lol

16. Dec 13, 2006

### Kurdt

Staff Emeritus
Argh what am I doing I'm just confirming your calc. What of course will be the answer is that the new pressure will be 1/4 of the original.

Apologies its 2am where I am.

17. Dec 13, 2006

### Kurdt

Staff Emeritus
To be fair even the initial conditions do not work in the equation are you sure you copied the question correctly. Thats what confused me.

Last edited: Dec 13, 2006
18. Dec 13, 2006

### Gokul43201

Staff Emeritus
This question is wrong. Any fixed amount of ideal gas at equilibrium, only has two independent degrees of freedom. Specifying any 2 of the 3 (P,V, T) quantites fixes the third one (through the ideal gas equation). For instance, a mole of gas at 220kPa occupying 1.2 x 10^-3 m^3 will have a temperature of about 30K (not 350K).

Be careful with the units. The value you used (0.08921) comes in units of L-atm/K-mol, while the SI value (8.314) has units of J/K-mol or Pa-m^3/K-mol.

19. Dec 13, 2006

### chrispsu

oh wow sorry i had no idea this went to 2 pages lol just thought there was no response

anyhow quartering the initial pressure worked! thanks!
and yea i copied and pasted it straight from the hwk since it is online. Many people have complained about this question to the professor so hopefully he will address it. Thanks though!