# Volume of a Balloon

1. May 7, 2010

### jagged06

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

Imagine that 9.9 g of liquid helium, initially at 4.20 K, evaporate into an empty balloon that is kept at 1.00-atm pressure. What is the volume of the ballonat the following?

(a) 23.0 K

(b) 299 K

2. Relevant equations

PV=nRT

3. The attempt at a solution

I converted th 9.9g of Helium to number of moles. n=2.473mol
and used that to find the volume at T=23 and T=299, but it isn't right.

steps
1) n=9.9g=(9.9g/(4.003g/mol))=2.47314 mol
R=8.314472 <-- gas constant
P= 1 as stated in the problem

V@ T=23 : V=(nRT)/P -> (2.47314*8.314472*23)/1 --> V=472.9466 L

V@ T=299 : V=(nRT)/P -> (2.47314*8.314472*299)/1 --> V=6148.30591237 L

I'm guessing I need to do something with T=4.2 first, but I'm not sure what.

Last edited: May 7, 2010
2. May 7, 2010

### kuruman

Please show your work, i.e. exactly what you did, and then perhaps we can determine where you went wrong.

3. May 7, 2010

### jagged06

edited to show work

4. May 7, 2010

### kuruman

Your problem is a mixed bag of units. When you write pV = nRT
p is measured in Pa (or N/m2) not atmospheres. How many Pa is one atmosphere? Your textbook should have the number. When you put in the correct pressure, the answer should come out in m3. If you want liters you need to make another conversion.