November 01, 2024, 02:22:00 AM
Forum Rules: Read This Before Posting


Topic: Combined Gas Law  (Read 3116 times)

0 Members and 1 Guest are viewing this topic.

Offline dannysmith

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Combined Gas Law
« on: February 26, 2009, 10:53:22 PM »
Hi everyone,

I have a question on combined Gas Laws (Charles+Boyle).  I have attempted this but no luck so far.

A balloon is filled with helium to a volume of 4.0 L when the pressure is 1.0 atm and the temperature is 27 C. It escapes and rises until the pressure is .25 atm and the temp is -23 C.  What is the new volume?

I started with Charles law because for Boyle the temperature must be constant.

Volume = Constant x Temp

4.0 L = constant x 300 K
4.0 L = 1/75 x 300 K

Next I plugged the constant in to new temp after gas escapes

?L= 1/75 x 250 K
?L= 3.3

Unfortunately I must be doing something wrong, because the answer key says the answer is 6.7 .

If anyone could help me, I would be so grateful!

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27820
  • Mole Snacks: +1808/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Combined Gas Law
« Reply #1 on: February 27, 2009, 03:35:57 AM »
Write combined gas law formula.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Sponsored Links