Volume of Balloon at 23K and 299K

[jagged06]

Homework Statement

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

Homework Equations

PV=nRT

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.

 

[kuruman]

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

 

[jagged06]

edited to show work

 

[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.