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

1. Copyright Sautter 2015 1

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
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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 …….
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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
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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
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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.
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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
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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
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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
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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
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