Posted by Anna on .
A vessel containing 39.5 cm^3 of helium gas at 25 degrees celsius and 106 kPa was inverted and placed in cold ethanol. As the gas contracted, ethanol was forced into the vessel to maintain the same pressure of helium. If tihs required 7.7 cm^3 of ethanol, what was the final temperature of the helium?

chem 
Trig,
PV/nT = PV/nT
UNITS:

P: atm (convert kPa to atm)
V: L (convert cm^3 to L)
n: mol (moles don't matter in this problem)
T: K (convert C to K)
( They need to fit with the ideal gas constant [R]'s units: [L*atm/K*mol] ) 
chem 
Trig,
(Pressure doesn't matter here either, since it remains constant. So you're left with V/T=V/T, which is Charles' Law.)