Transcript SOLVING GAS LAW PROBLEMS
Copyright Sautter 2003
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
GAS LAW FORMULAE
• BOYLE’S LAW: P 1 x V 1 = P 2 x V 2 • • • • • CHARLES LAW: V 1 / T 1 = V 2 / T 2 GAY-LUSSAC’S LAW: P 1 / T 1 = P 2 / T 2
KELVIN = 273 + DEGREES CELSIUS
COMBINED GAS LAW (P 1 x V 1 ) / T 1 = (P 2 DALTON’S LAW x V 2 ) / T 2 P TOTAL = P GAS A + P GAS B + P GAS C + P …….
GAS LAW FORMULAE (CONT’D)
• DALTON’S LAW (CONT’D) • P GAS A = (N GAS A / N TOTAL ) x P TOTAL • 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
GAS LAW FORMULAE (CONT’D)
• GRAHAM’S LAW OF DIFFUSION v 2 / v 1 = ( m 1 / m 2 ) 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
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: P 1 x V 1 = P 2 x V 2 , P 1 = 2.0 ATM V 1 = 500 ML, P 2 = 2090 MM / 760 = 2.75 ATM V 2 = ( P 1 x V 1 ) / P 2 V 2 = (2.0 x 500) / 2.75 = 364 ML NOTE: BOYLE’S LAW IS INVERSE, AS PRESSURE INCREASES, VOLUME DECREASES.
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: P 1 x V 1 = P 2 x V 2 , P 1 = 1140 MM V 1 = 6.0 L, V 2 = 2000 ML / 1000 = 2.0 L P 2 = ( P 1 x V 1 ) / V 2 P 2 = (1140 x 6.0) / 2.0 = 3420 MM OF HG
SOLVING CHARLES LAW PROBLEMS
• WHAT IS THE VOLUME OF HYDROGEN WHEN 300 ML ARE HEATED FROM 35 CELSIUS TO 80 CELSIUS ?
• SOLUTION: V 1 / T 1 = V 2 / T 2 , V 1 = 300 ML •
KELVIN = 273 + DEGREES CELSIUS
• T 1 = (35 +273) = 308 K, T 2 = (80 + 273) = 353 K • V 2 = (V 1 x T 2 ) / T 1 • V 2 = (300 x 353) / 308 = 344 ml
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: V 1 / T 1 = V 2 / T 2 , V 1 = 500 ML, V T 1 = 300 K 2 = 350 ML • T 2 = (V 2 x T 1 ) / V 1 • T 2 = (350 x 300) / 500 = 210 K •
KELVIN = 273 + DEGREES CELSIUS
•
210 = 273 + C 0 , C 0 = - 63
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
C 0
?
SOLUTION: P 1 / T 1 = P 2 / T 2 , P 1 = 3.0 ATM, T 1 = 200 K T 2 = 1000
C 0
• •
KELVIN = 273 + DEGREES CELSIUS K = 273 + 1000 = 1273 K
• P 2 = (P 1 x T 2 ) / T 1 • P 2 = (3.0 x 1273) / 200 = 19.1 ATM
• •
SOLVING GAY-LUSSAC LAW PROBLEMS
A SAMPLE OF CHLORINE IS RAISED TO 1140 MM OF HG FROM A PRESSURE OF 0.50 ATM. IF THE ORIGINAL TEMPERATURE WAS 500 K, WHAT IS THE NEW TEMPERATURE IN CELSIUS ?
SOLUTION: P 1 / T 1 = P 2 / T 2 , P 1 = 0.50 ATM, P 2 = 1140 / 760 = 1.5 ATM T 1 = 500 K T 2 = (P 2 x T 1 ) / P 1 T 2 = (1.5 x 500) / 0.50 = 1500 K • •
KELVIN = 273 + DEGREES CELSIUS 1500 = 273 + C 0 , C 0 = 1227
SOLVING COMBINED LAW PROBLEMS
• • • WHAT IS THE NEW VOLUME OF 650 ML OF NITROGEN AT 273 K AND 2.0 ATMS WHEN IT IS HEATED TO 819 K AND REDUCED TO 1.0 ATM PRESSURE ?
SOLUTION: (P 1 x V 1 ) / T 1 = (P 2 x V 2 ) / T 2 , P 1 = 2.0 ATM, P 2 = 1.0 ATM V 2 = (P 1 x V 1 x T 2 ) / (P 2 x T 1 ) T 1 = 273 K, T 2 = 819 K V 1 = 650 ml V 2 = (2.0 x 650 x 819) / (1.0 x 273) = 3900 ml
SOLVING COMBINED LAW PROBLEMS
• • • WHAT IS THE NEW PRESSURE OF 850 ML OF ARGON AT 1092 K AND 5.0 ATMS WHEN IT IS COOLED TO 273 C AND REDUCED TO A VOLUME OF 300 ML ?
SOLUTION: (P 1 x V 1 ) / T 1 = (P 2 x V 2 ) / T 2 , P 1 = 5.0 ATM, V 2 = 300 ML P 2 = (P 1 x V 1 x T 2 ) / (V 2 x T 1 ) T 1 = 1092 K, V 1 = 850 ML T 2 = (273 + 273) = 546 K P 2 = (5.0 x 850 x 546) / (300 x 1092) = 7.1 ATM
SOLVING DALTON’S LAW PROBLEMS
P A TANK CONTAINS THREE GASES, N 2 , Cl 2 AND O 2 . THE NITROGEN PRESSURE IS 2.0 ATM, THE CHLORINE 380 MM OF HG AND THE OXYGEN 5.0 ATM. WHAT IS THE PRESSURE IN THE TANK ?
TOTAL = P GAS A + P GAS B + P GAS C + P …….
P TOTAL = 2.0 ATM + (380/760)ATM + 5.0 ATM = 7.5 ATM OR 7.5 x 760 = 5700 MM OF HG
SOLVING DALTON’S LAW PROBLEMS
• • • • A TANK CONTAINS THREE GASES, H 2 , Br 2 AND O 2 . THE MASS OF HYDROGEN IS 2.0 GRAMS, THE BROMINE MASS IS 240 GRAMS AND THE OXYGEN MASS IS 16.0 GRAMS. THE TOTAL PRESSURE IN THE TANK WAS 4.0 ATM. WHAT IS THE PRESSURE OF THE OXYGEN IN THE TANK ?
SOLUTION: P GAS A = (N GAS A / N TOTAL ) x P TOTAL MOLES = GRAMS / MOLAR MASS H 2 = 2.0 / 2.0 = 1.0 MOLES, Br 2 = 240 / 160 =1.5 MOLES O 2 = 16.0 / 32 = 0.50 MOLES P O2 = ( (0.50) / (1.0 + 1.5 + 0.50)) x 4.0 = 0.67 ATM
SOLVING IDEAL GAS LAW PROBLEMS
WHAT IS THE TEMPERATURE OF 68.0 GRAMS OF OF HYDROGEN SULFIDE GAS WITH A VOLUME OF 6.0 LITERS AND A PRESSURE OF 5.0 ATMS ?
SOLUTION: P x V = N x R x T, P = 5.0 ATM, V = 6.0 LITERS N = 68.0 / 34.0 = 2.0 MOLES R = 0.0821 ATM x L / MOLES x K • • T = (P x V) / (N x R) T = (5.0 x 6.0) / (2.0 x 0.0821) = 183 K
KELVIN = 273 + DEGREES CELSIUS 183 = 273 + C 0 , C 0 = -90
SOLVING IDEAL GAS LAW PROBLEMS
WHAT IS THE VOLUME OF 64.0 GRAMS OF OF OXYGEN GAS WITH A TEMPERATURE OF 25 DEGREES CELSIUS AND A PRESSURE OF 3.0 ATMS ?
SOLUTION: P x V = N x R x T,
KELVIN = 273 + DEGREES
T= 25 + 273 = 298 K, P = 3.0 ATM N = 64.0 / 32.0 = 2.0 MOLES R = 0.0821 ATM x L / MOLES x K • V = (N x R x T) / P V = (2.0 x 0.0821 x 298) / 3.0
183 = 273 + C 0 , C 0 = -90
SOLVING GRAHAM’S LAW PROBLEMS
THE AVERAGE MOLECULAR SPEED OF AN OXYGEN MOLECULE AT A SPECIFIC TEMPERATURE IS 500 M/SEC. WHAT IS THE AVERAGE SPEED OF A NEON MOLECULE AT THE SAME TEMPERATURE ? v 2 / v 1 = ( m 1 / m 2 ) 1/2 MOLAR MASS OXYGEN = 32.0, MOLAR MASS NEON = 20.0
m 1 = 32, m 2 = 20.0, v 1 = 500 M / SEC v 2 = ( m 1 / m 2 ) 1/2 (v 1 ) = (32.0/ 20.0) 1/2 (500) =632 M/SEC
SOLVING GRAHAM’S LAW PROBLEMS
WHAT IS MOLECULAR MASS OF SUBSTANCE X, IF AT A SPECIFIC TEMPERATURE THE SPEED OF A NITROGEN MOLECULE IS 800 M/SEC. THE AVERAGE SPEED OF THE UNKNOWN GAS IS 650 M/SEC AT THE SAME TEMPERATURE ? v 2 / v 1 = ( m 1 / m 2 ) 1/2 (squaring both sides gives) m 1 / m 2 = v 2 2 / v 1 2 ( solving for m 1 gives) m 1 = (v 2 2 / v 1 2 ) x m 2 m 2 v 2 = Molar mass of Nitrogen = 28.0 = velocity of Nitrogen = 800 m / s v 1 = velocity of unknown gas = 650 m /s m 1 = (800 2 / 650 2 ) x 28 = 42.4
SOLVING STP PROBLEMS
ONE MOLE OF ANY GAS AT STP ( 0 C OR 273 K AND 1 ATM OR 760 MM OF HG) OCCUPIES 22.4 LITERS HOW MANY GRAMS OF HYDROGEN ARE CONTAINED IN 89.6 LITERS AT STP ?
SOLUTION :
Liters divide by 22.4 moles 89.6 / 22.4 = 4.0 MOLES MOLES x MOLAR MASS = GRAMS 4.0 x 2.0 = 8.0 GRAMS HYDROGEN
SOLVING STP PROBLEMS
MOLE OF ANY GAS AT STP ( 0 C OR 273 K AND 1 ATM OR 760 MM OF HG) OCCUPIES 22.4 LITERS WHAT IS THE VOLUME OF 96.0 GRAMS OF SULFUR DIOXIDE AT STP ?
SOLUTION :
GRAMS / MOLAR MASS = MOLES 96.0 / 64.0 = 1.5 MOLES MOLES X 22.4 LITER s 1.5 x 22.4 = 33.6 LITERS