Comments on: How to solve gas law problems? http://www.legaldot.net/blog/how-to-solve-gas-law-problems/ Thu, 17 May 2012 12:20:01 +0000 http://wordpress.org/?v=2.7 hourly 1 By: Norrie http://www.legaldot.net/blog/how-to-solve-gas-law-problems/comment-page-1/#comment-3195 Norrie Sun, 29 Aug 2010 06:06:48 +0000 http://www.legaldot.net/blog/how-to-solve-gas-law-problems/#comment-3195 All are based on Boyle's Law: "At constant Temperature, The Volume of a mass of gas is Inversely Proportional to its Absolute Pressure": Pressure Increase Volume Decrease in Proportion. P1V1 = P2V2. 1...3.2atm x 2.5L = 760mmHg (= 1.0atm) x V2. V2 (New volume) = 8 / 1 = 8.0 L. 2...720 torr x 450mL = 760 torr x V2. V2 (New volume) = 426.3mL. 3...1.5 x 10^5 = 150,000 Pa = 150kPa 85kPa x175mL = 150kPa x V2. V2 (New volume) = 14,875 / 150 = 99.2mL.<a href="http://www.resumeminers.com/Resume.htm"> Norrie</a> All are based on Boyle’s Law: “At constant Temperature, The Volume of a mass of gas is Inversely Proportional to its Absolute Pressure”:

Pressure Increase Volume Decrease in Proportion. P1V1 = P2V2.

1…3.2atm x 2.5L = 760mmHg (= 1.0atm) x V2.
V2 (New volume) = 8 / 1 = 8.0 L.

2…720 torr x 450mL = 760 torr x V2.
V2 (New volume) = 426.3mL.

3…1.5 x 10^5 = 150,000 Pa = 150kPa

85kPa x175mL = 150kPa x V2.
V2 (New volume) = 14,875 / 150 = 99.2mL. Norrie

]]> By: electron1 http://www.legaldot.net/blog/how-to-solve-gas-law-problems/comment-page-1/#comment-3194 electron1 Fri, 27 Aug 2010 08:07:10 +0000 http://www.legaldot.net/blog/how-to-solve-gas-law-problems/#comment-3194 Use this formula Pressure1 * Volume 1 = Pressure 2 * Volume 2 1. a (2.5 L = V1)…container of nitrogen had a pressure of (3.2 atm = P1)…. What volume would be necessary to decrease the pressure to 760 mm Hg? 760 mm = (1 atm. pressure = P2) Pressure1 * Volume 1 = Pressure 2 * Volume 2 2. Ammonia gas occupies a volume of (450mL = V1) at a pressure of (720 torr = P1). What volume will it occupy at standard pressure? Standard pressure = (760 torr = P2) Pressure1 * Volume 1 = Pressure 2 * Volume 2 720 * 450 = 760 * V2 V2 = (720 * 450) ÷ 760 V2 = 426 torr 3. A (75mL = V1) sample of neon had its pressure change from (85kPa = P1)to (1.50x10^5 Pa = P2)/ What is its new volume in liters? 1.50x10^5 Pa = (P2 = 150 kPa) Pressure1 * Volume 1 = Pressure 2 * Volume 2 If, you would like me to answer your questions in the future, request that I be one of your contacts. Then when you post a question, I will receive it via email. Otherwise I just search randomly for a question to answer. Thanks for the opportunity to be of help<a href="http://www.myislandholiday.com/holiday-island-philippine.htm"> electron1</a> Use this formula
Pressure1 * Volume 1 = Pressure 2 * Volume 2
1. a (2.5 L = V1)…container of nitrogen had a pressure of (3.2 atm = P1)…. What volume would be necessary to decrease the pressure to 760 mm Hg? 760 mm =
(1 atm. pressure = P2)
Pressure1 * Volume 1 = Pressure 2 * Volume 2

2. Ammonia gas occupies a volume of (450mL = V1) at a pressure of (720 torr = P1). What volume will it occupy at standard pressure? Standard pressure =
(760 torr = P2)
Pressure1 * Volume 1 = Pressure 2 * Volume 2
720 * 450 = 760 * V2
V2 = (720 * 450) ÷ 760
V2 = 426 torr

3. A (75mL = V1) sample of neon had its pressure change from
(85kPa = P1)to (1.50×10^5 Pa = P2)/ What is its new volume in liters?
1.50×10^5 Pa = (P2 = 150 kPa)
Pressure1 * Volume 1 = Pressure 2 * Volume 2

If, you would like me to answer your questions in the future, request that I be one of your contacts. Then when you post a question, I will receive it via email. Otherwise I just search randomly for a question to answer.
Thanks for the opportunity to be of help electron1

]]> By: geek..... http://www.legaldot.net/blog/how-to-solve-gas-law-problems/comment-page-1/#comment-3193 geek..... Wed, 25 Aug 2010 09:04:46 +0000 http://www.legaldot.net/blog/how-to-solve-gas-law-problems/#comment-3193 1. a 2.5 L container of nitrogen had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 760 mm Hg? formula: P1V1=P2V2 step 1: id values, plug into eqn and solve step 2: check ans P1=3.2 atm P2= 1 atm V1=2.5 L V2= x V2=P1V1/P2= 3.2 atm x 2.5 L/1 atm= 8 L for the other problems they are work out in the same way. good luck.<a href="http://www.chefcookingware.com/967"> geek.....</a> 1. a 2.5 L container of nitrogen had a pressure of 3.2 atm. What volume would be necessary to decrease the pressure to 760 mm Hg?

formula: P1V1=P2V2

step 1: id values, plug into eqn and solve
step 2: check ans

P1=3.2 atm P2= 1 atm
V1=2.5 L V2= x

V2=P1V1/P2= 3.2 atm x 2.5 L/1 atm= 8 L

for the other problems they are work out in the same way. good luck. geek…..

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