Solving Gas Laws

15,556 views

Published on

How to solve chemistry problems using Boyles Law, Charles Law, Gay Lussac's Law, the Combined Gas Law, Avogadro's Hypothesis, and Dalton's Law.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f

Published in: Education, Business, Technology
1 Comment
5 Likes
Statistics
Notes
  • ALSO
    I have written six books
    'Sticks - A Golfer’s Tale' - A hacker’s dream comes true.
    'Fish Farm' - Revenge of the old people.
    'Coach' - A mystery in Old school football town in a rural, bigoted, fifties town.
    'The Three Dollar Phoenix' - A mystery set in Newark, New Jersey in the 1970s.
    'The Divine Comedy MMIX' - A humorous play about Jesus returning.
    'The Blood of Judas' - A horror story of revenge set in Nazi Germany.
    All are available at www.smashwords.com
    I have video trailers for 'Coach', 'Fish Farm' and 'The Blood of Judas' at:
    http://www.youtube.com/watch?v=xXSD5Kz-fDY
    http://www.youtube.com/watch?v=a9PTRb14ldc
    http://www.youtube.com/watch?v=ToPp9k9Oq-o
    http://www.youtube.com/watch?v=3eBhMZbsP-I
    Please take a look. Thanks.
    Walt Sautter - wsautter@optonline.net
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
No Downloads
Views
Total views
15,556
On SlideShare
0
From Embeds
0
Number of Embeds
34
Actions
Shares
0
Downloads
310
Comments
1
Likes
5
Embeds 0
No embeds

No notes for slide

Solving Gas Laws

  1. 1. Copyright Sautter 2015 1
  2. 2. SOLVING GAS LAW PROBLEMS • BOYLE’S LAW • CHARLES LAW • GAY-LUSSAC’S LAW • THE COMBINED GAS LAW • THE IDEAL GAS LAW • DALTON’S LAW • GRAHAM’S LAW OF DIFFUSION 2
  3. 3. GAS LAW FORMULAE • BOYLE’S LAW: P1 x V1 = P2 x V2 • CHARLES LAW: V1 / T1 = V2 / T2 • GAY-LUSSAC’S LAW: P1 / T1 = P2 / T2 • KELVIN = 273 + DEGREES CELSIUS • COMBINED GAS LAW (P1 x V1 ) / T1 = (P2 x V2 ) / T2 • DALTON’S LAW P TOTAL = PGAS A + P GAS B + P GAS C + P ……. 3
  4. 4. GAS LAW FORMULAE (CONT’D) • DALTON’S LAW (CONT’D) • PGAS A = (N GAS A / N TOTAL) x PTOTAL • AVOGADRO’S HYPOTHESIS “EQUAL VOLUMES OF DIFFERENT GASES, AT THE SAME TEMPERATURE AND PRESSURE, CONTAIN EQUAL MOLES” UNIVERSAL GAS LAW P x V = N x R x T 4
  5. 5. GAS LAW FORMULAE (CONT’D) • GRAHAM’S LAW OF DIFFUSION v2 / v1 = ( m1 / m2)1/2 v = average molecular velocity m = molecular mass ONE MOLE OF ANY GAS OCCUPIES 22.4 LITERS AT STP CONDITIONS Liters divide by 22.4 moles Liters multiply by 22.4 moles 5
  6. 6. SOLVING BOYLE’S LAW PROBLEMS • WHAT IS THE VOLUME OF 500 ML OF NEON GAS AT 2.0 ATMS OF PRESSURE WHEN ITS PRESSURE IS CHANGED TO 2090 MM OF HG ? • SOLUTION: P1 x V1 = P2 x V2 , P1 = 2.0 ATM V1 = 500 ML, P2 = 2090 MM / 760 = 2.75 ATM V2 = ( P1 x V1) / P2 V2 = (2.0 x 500) / 2.75 = 364 ML NOTE: BOYLE’S LAW IS INVERSE, AS PRESSURE INCREASES, VOLUME DECREASES. 6
  7. 7. SOLVING BOYLE’S LAW PROBLEMS • IF 6.0 LITERS OF OXYGEN AT 1140 MM OF HG IS REDUCED TO A VOLUME OF 2000 ML, WHAT IS THE NEW PRESSURE OF THE GAS ? • SOLUTION: P1 x V1 = P2 x V2 , P1 = 1140 MM V1 = 6.0 L, V2 = 2000 ML / 1000 = 2.0 L P2 = ( P1 x V1) / V2 P2 = (1140 x 6.0) / 2.0 = 3420 MM OF HG 7
  8. 8. SOLVING CHARLES LAW PROBLEMS • WHAT IS THE VOLUME OF HYDROGEN WHEN 300 ML ARE HEATED FROM 35 CELSIUS TO 80 CELSIUS ? • SOLUTION: V1 / T1 = V2 / T2 , V1 = 300 ML • KELVIN = 273 + DEGREES CELSIUS • T1 = (35 +273) = 308 K, T2 = (80 + 273) = 353 K • V2 = (V1 x T2 ) / T1 • V2 = (300 x 353) / 308 = 344 ml 8
  9. 9. SOLVING CHARLES LAW PROBLEMS • A 500 ml sample of carbon dioxide is reduced to 350 ml by cooling. If the original temperature has 300 K, what is the new temperature in degrees Celsius ? • SOLUTION: V1 / T1 = V2 / T2 , V1 = 500 ML, V2 = 350 ML T1 = 300 K • T2 = (V2 x T1 ) / V1 • T2 = (350 x 300) / 500 = 210 K • KELVIN = 273 + DEGREES CELSIUS • 210 = 273 + C0 , C0 = - 63 9
  10. 10. SOLVING GAY-LUSSAC LAW PROBLEMS • WHAT IS THE PRESSURE OF A CONFINED GAS WITH AN ORIGINAL PRESSURE OF 3.0 ATM AND A TEMPERATURE OF 200K IF THE TEMPERATURE IS INCREASED TO 1000 C0 ? • SOLUTION: P1 / T1 = P2 / T2 , P1 = 3.0 ATM, T1 = 200 K T2 = 1000 C0 • KELVIN = 273 + DEGREES CELSIUS • K = 273 + 1000 = 1273 K • P2 = (P1 x T2 ) / T1 • P2 = (3.0 x 1273) / 200 = 19.1 ATM 10
  11. 11. 11 Click Here

×