A husband buys a helium-filled anniversary balloon for his wife. The balloon has a volume of 4.4 L in the warm store at 74 degree F. When he takes it outside, where the temperature is 55 degree F, he finds it has shrunk. By how much has the volume decreased?

Question

This is what I did:

74-32=45(5/9)=23+273=296K
55-32=45(5/9)=12.8+273=286K

4.4/296 = V2/286
286(4.4)/296 = 4.25L

But I'm getting the wrong answer. What am I doing wrong? Help please.

Answer 1

To convert Fahrenheit to Rankine, add 460.

74 F = 534 R = 297 K

55 F = 515 R = 286 K

If pressure stayed the same, V/T remains constant and

V2 = (286/297)4.4 = 4.24 L

Actually, a falling volume decreases the pressure in an already-stretched balloon. One would need more data on balloon tension and elasticity to take this into account.

Answer 2

This may help.
http://www.digipac.ca/chemical/gaslaws/TchrNotesPVB.htm