1. Solving
Gas Law
Problems
Copyright Sautter 2003

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

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 = 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 …….

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

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 moles
Liters multiply by moles

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 = MM / 760 = 2.75 ATM
V2 = ( P1 x V1) / P2
V2 = (2.0 x 500) / = 364 ML
NOTE: BOYLE’S LAW IS INVERSE, AS PRESSURE INCREASES, VOLUME DECREASES.

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 = ML / 1000 = 2.0 L
P2 = ( P1 x V1) / V2
P2 = (1140 x 6.0) / 2.0 = MM OF HG

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 = DEGREES CELSIUS
T1 = ( ) = 308 K, T2 = ( ) = 353 K
V2 = (V1 x T2 ) / T1
V2 = (300 x 353) / 308 = 344 ml

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 = DEGREES CELSIUS
210 = C0 , C0 = - 63

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 = C0
KELVIN = DEGREES CELSIUS
K = = K
P2 = (P1 x T2 ) / T1
P2 = (3.0 x 1273) / 200 = ATM

11. 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:
P1 / T1 = P2 / T2 , P1 = 0.50 ATM, P2 = 1140 / 760 = 1.5 ATM
T1 = 500 K
T2 = (P2 x T1 ) / P1
T2 = (1.5 x 500) / 0.50 = K
KELVIN = DEGREES CELSIUS
1500 = C0 , C0 = 1227

12. 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:
(P1 x V1 ) / T1 = (P2 x V2 ) / T2, P1 = 2.0 ATM, P2 = 1.0 ATM
T1 = 273 K, T2 = 819 K
V1 = 650 ml
V2 = (P1 x V1 x T2 ) / (P2 x T1 )
V2 = (2.0 x x 819) / (1.0 x 273) = ml

13. 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:
(P1 x V1 ) / T1 = (P2 x V2 ) / T2, P1 = 5.0 ATM, V2 = 300 ML
T1 = K, V1 = 850 ML
T2 = ( ) = 546 K
P2 = (P1 x V1 x T2 ) / (V2 x T1 )
P2 = (5.0 x x 546) / (300 x 1092) = 7.1 ATM

14. SOLVING DALTON’S LAW PROBLEMS
A TANK CONTAINS THREE GASES, N2 , Cl2 AND O2 . 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 ?
P TOTAL = PGAS A P GAS B P GAS C + P …….
P TOTAL = 2.0 ATM + (380/760)ATM ATM = 7.5 ATM
OR x = 5700 MM OF HG

15. SOLVING DALTON’S LAW PROBLEMS
A TANK CONTAINS THREE GASES, H2 , Br2 AND O2 . 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:
PGAS A = (N GAS A / N TOTAL) x PTOTAL
MOLES = GRAMS / MOLAR MASS
H2 = 2.0 / 2.0 = 1.0 MOLES, Br2 = 240 / 160 =1.5 MOLES
O2 = 16.0 / 32 = 0.50 MOLES
PO2 = ( (0.50) / ( )) x = ATM

16. 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 / = 2.0 MOLES
R = ATM x L / MOLES x K
T = (P x V) / (N x R)
T = (5.0 x 6.0) / (2.0 x ) = 183 K
KELVIN = DEGREES CELSIUS
= C0 , C0 = -90

17. 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 = DEGREES
T= = 298 K, P = 3.0 ATM
N = 64.0 / = 2.0 MOLES
R = ATM x L / MOLES x K
V = (N x R x T) / P
V = (2.0 x x 298) / 3.0
= C0 , C0 = -90

18. 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 ?
v2 / v1 = ( m1 / m2)1/2
MOLAR MASS OXYGEN = 32.0, MOLAR MASS NEON = 20.0
m1 = 32, m2 = 20.0, v1 = 500 M / SEC
v2 = ( m1 / m2)1/2 (v1) = (32.0/ 20.0) 1/2 (500) =632 M/SEC

19. 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 ?
v2 / v1 = ( m1 / m2)1/2 (squaring both sides gives)
m1 / m2 = v22 / v12 (solving for m1 gives)
m1 = (v22 / v12) x m2
m2 = Molar mass of Nitrogen = 28.0
v2 = velocity of Nitrogen = 800 m / s
v1 = velocity of unknown gas = 650 m /s
m1 = (8002 / 6502) x 28 = 42.4

20. SOLVING STP PROBLEMS Liters divide by 22.4 moles
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 moles
89.6 / = 4.0 MOLES
MOLES x MOLAR MASS = GRAMS
4.0 x = 8.0 GRAMS HYDROGEN

21. SOLVING STP PROBLEMS 1.5 x 22.4 = 33.6 LITERS
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 / = 1.5 MOLES
MOLES X LITERs
1.5 x = LITERS

22. THE END