You have a balloon containing 1 L of air at STP in a vacuum chamber. What will the volume of the balloon be when you reduce the pressure by half and increase the temperature to 373 K? (Use the combined gas law PV/T = k or P1V1/T1 = P2V2/T2.)

Question

daniellelovee · Answer

2.73 L

ichliebedich · Answer

This question is just an ideal gas equation, or PV=nRT. 
You are given that the gas originally is 1 L, and it is at STP, which is P=1 atm, T=273 K. At STP R=.0821, so what you do first is figure out how many moles (n) are in the gas. 
(1 atm)(1 L) = n (.0821 Latm/Kmol) (273 K) 
n = .0446 mol 
Now, you have to figure out what volume that number of mols will take up in the new conditions. Use the ideal gas law again, this time solving for volume. 
(.5 atm) V = (.0446 mol) (.0821 Latm/Kmol) (373 K) 
V = 2.731 L

daniellelovee · Answer

ok you are being of great help danke :p

ichliebedich · Answer

bitte