377
GAS LAW WORKSHEET 1 KEY
1. A sample of oxygen gas occupies a volume of 436.2 mL at 1.02 atm. If the temperature is held constant,
what would the pressure of this gas be when the gas is compressed to 231.6 mL?
P1 V1
PV
= 2 2
T1
T2
since T1 = T2
P1 V1 = P2 V2
P2 = P1
V1
436.2 mL 
= 1.02 atm 
 = 1.92 atm
V2
 231.6 mL 
To check with reasoning: decreased V leads to increased P, so the volume ratio should be greater than
one to yield the expected increased pressure.
2. If a gas originally occupying 6.75 L at 19.21 ºC and 762.5 torr is compressed to give a pressure of
1.26 atm at 26.35 ºC, what would the new volume be?
P1 V1
PV
= 2 2
T1
T2
 T  P 
299.50 K   762.5 torr   1 atm 
V2 = V1  2  1  = 6.75 L 


 = 5.51 L
 292.36 K   1.26 atm   760 torr 
 T1  P2 
To check with reasoning: increased T leads to increased V, so the temperature ratio should be greater
than one to yield the expected increase in volume. Increased P leads to decreased V, so the pressure ratio
should be less than one to yield the expected decreased volume.
3. Calculate the number of grams of hydrogen sulfide gas, H2S, in a 3.6 L container at 32.6 ºC and 712
mmHg.
PV =
g
RT
M
g=
PVM
=
RT
g
mol  1 atm  = 4.6 g H S


2
L atm
0.082058
(305.8 K )  760 mmHg 
K mol
712 mmHg (3.6 L ) 34.080
4. Calculate the density of 0.625 g of carbon dioxide at 26.32 ºC and 1.03 atm.
PV =
g
RT
M
g 

1.03 atm  44.010

g
PM
mol 

=
=
= 1.84 g/L CO 2
L atm
V
RT
0.082058
299.47
K
(
)
K mol
378
Chapter 10 Worksheet Keys
CHEMISTRY 151 – COMBINED GAS LAW KEY
1. A McLeod gauge is an instrument used to measure extremely low pressures. Assume that a 250.0 mL
sample of gas from a low pressure system is compressed in a McLeod gauge to a volume of 0.0525 mL,
where the pressure of the sample is 0.0355 atm. What was the original pressure of the gas in the system?
V1 = 250.0 mL V2 = 0.0525 mL
P1 V1 = P2 V2
P1 =
P2 = 0.0355 atm
P1 = ?
P2 V2
 0.0525 mL 
−6
= 0.0355 atm 
 = 7.46 x 10 atm
V1
 250.0 mL 
2. A balloon containing 5.0 dm3 of gas at 14 °C and 100.0 kPa rises to an altitude of 2000 m, where the
temperature is 20 °C. The pressure of the gas in the balloon is now 79.0 kPa. What is the volume of gas
in the balloon at this altitude?
V1 = 5.0 dm3
T1 = 14 ºC + 273.15 = 287 K
T2 = 20 ºC + 273.15 = 293 K
P1 = 100.0 kPa
P2 = 79.0 kPa
V2 = ?
 T  P 
 293 K  100.0 kPa 
3
V2 = V1  2  1  = 5.0 dm3 

 = 6.5 dm
T
P
287
K
79.0
kPa



 1  2 
P1 V1
PV
= 2 2
T1
T2
3. A cylinder contains carbon dioxide gas at 6.50 atm. Enough carbon dioxide gas escapes from this cylinder
into a 24.8 L container to bring the pressure of this container to 1.05 atm at 21 °C. If the volume of the
container then changes, what would the new volume of gas be if its pressure goes to 1.28 atm at 30 °C?
V1 = 24.8 L
T1 = 21 ºC + 273.15 = 294 K
P1 = 1.05 atm
P2 = 1.28 atm
T2 = 30 ºC + 273.15 = 303 K V2 = ?
P1 V1
PV
= 2 2
T1
T2
 T  P 
 303 K   1.05 atm  21.0 L
V2 = V1  2  1  = 24.8 L 
=

 294 K   1.28 atm 
 T1  P2 
379
CHEMISTRY 151 - IDEAL GAS EQUATION KEY
1. The maximum safe pressure that a certain 4.00 L vessel can hold is 3.50 atm. If the vessel contains
0.410 mole of gas, what is the maximum temperature to which the vessel can be subjected?
PV = nRT
T=
3.50 atm (4.00 L )
PV
=
= 416 K or 143 °C
L atm 
nR
0.410 mol  0.082058

K mol 

2. A barge containing 640 tons of liquid chlorine had an accident on the Ohio River. What volume would
this amount of chlorine occupy if it escaped into the atmosphere as a gas at 756 mmHg and 15 °C?
PV =
g
RT
M
gRT
V=
=
PM
L atm 
640 tons  0.082058
 288 K 2000 lb 453.6 g 760 mmHg




K mol 

8


 = 1.95 x 10 L

g

 1 ton   1 lb   1 atm 
756 mmHg  70.906

mol 

3. What is the density of hydrogen gas at 299.7 K and 386.98 mmHg?
PV =
g
RT
M
g 
386.98 mmHg  2.01594

g
PM
mol   1 atm 

=
=

 = 0.04174 g/L
L atm
V
RT
760
mmHg


0.082058
( 299.7 K )
K mol
4. A radioactive metal atom decays by emitting an alpha particle, which is a helium nucleus. The alpha particles are collected as helium gas. A sample of helium with a volume of 12.05 mL was obtained at
765 mmHg and 23 °C. How many atoms decayed during the period of the experiment?
PV
RT
765 mmHg (12.05 mL )  1 atm   1 L   6.022 x 1023 atoms 
? atoms =


 3

L atm
1 mol

0.082058
(296 K )  760 mmHg   10 mL  
K mol
= 3.01 x 1020 atoms He
PV = nRT
n=
380
Chapter 10 Worksheet Keys
GAS STOICHIOMETRY KEY
1. In 1793 the French physicist Jacques Charles supervised and took part in the first human flight in a
hydrogen balloon. The hydrogen was produced by reacting iron filings with sulfuric acid to yield hydrogen gas and iron(II) sulfate. What volume of hydrogen gas at 22 °C and 756 mmHg will be formed from
the reaction of 500 gallons of 12 M sulfuric acid?
Fe(s) + H2SO4(aq)
→
FeSO4(aq) + H2( g)
 3.785 L   12 mol H2 SO4  1 mol H2   0.082058 L atm 
? L H2 = 500 gal H2 SO4 




K mol

 1 gal   1 L H2 SO4 soln  1 mol H2 SO4  
 295 K   760 mmHg 
5


 = 5.5 x 10 L H2
 756 mmHg   1 atm 
2. Oxygen gas can be generated by heating potassium chlorate to yield oxygen gas and potassium chloride.
If 85.0 g of potassium chlorate are reacted in a 2.50 L container and the oxygen gas is cooled to 21 °C in
the same container, what will the partial pressure of the oxygen be?
2KClO3(s)
→
2KCl(s) + 3O2(g )
 1 mol KClO3  3 mol O2 
? mol O2 = 85.0 g KClO3 

 = 1.04 mol O2
122.553
g
KClO
2
mol
KClO
3 
3 

L atm 
1.04 mol O2  0.082058
 294 K
nRT
K mol 

P=
=
= 10.0 atm
V
2.50 L
or you can do the calculation in one step.
 1 mol KClO3  3 mol O2   0.082058 L atm  294 K 
? atm = 85.0 g KClO3 



 = 10.0 atm
K mol
 2.50 L 
 122.553 g KClO3  2 mol KClO3  
3. When carbon disulfide is completely burned, it yields carbon dioxide gas and sulfur dioxide gas. The
sulfur dioxide can be removed from the gas mixture by bubbling the mixture through water to dissolve
the sulfur dioxide leaving the insoluble carbon dioxide.
CS2 + 3O2 → CO2 + 2SO2
a. What volume of carbon dioxide gas at 19.6 °C and 756.4 mmHg will form from the complete
combustion of 2.50 L of carbon disulfide gas at 19.6 °C and 756.4 mmHg?
 1 L CO2 
? L CO2 = 2.50 L CS2 
 = 2.50 L CO 2
 1 L CS2 
b. What volume of carbon dioxide gas at 19.6 °C and 756.4 mmHg will form from the complete combustion of 2.50 L of carbon disulfide gas at 18.1 °C and 4.75 atm?
K mol

 4.75 atm   1 mol CO2 
? L CO2 = 2.50 L CS2 



 0.082058 L atm   291.3 K   1 mol CS2 
 0.082058 L atm   292.8 K   760 mmHg  = 12.0 L CO




2
K mol

  756.4 mmHg   1 atm 
381
c. What volume of carbon dioxide gas at 19.6 °C and 756.4 mmHg will form from the complete combustion of 2.50 L of carbon disulfide gas at 18.1 °C and 4.75 atm with 3.25 L of oxygen at 18.1 °C
and 10.2 atm?
K mol

 4.75 atm   1 mol CO2 
? L CO2 = 2.50 L CS2 



 0.082058 L atm   291.3 K   1 mol CS2 
 0.082058 L atm   292.8 K   760 mmHg  = 12.0 L CO




2
K mol

  756.4 mmHg   1 atm 
K mol

 10.2 atm   1 mol CO2 
? L CO2 = 3.75 L O2 



 0.082058 L atm   291.3 K   3 mol O2 
 0.082058 L atm   292.8 K   760 mmHg  = 11.2 L CO




2
K mol

  756.4 mmHg   1 atm 
This is a limiting reactant problem. The CS2 would run out when 12.0 L CO2 are formed. The
O2 runs out when 11.2 L CO2 is formed. The O2 is the limiting reactant.
382
Chapter 10 Worksheet Keys
CHEMISTRY 151 - GAS STOICHIOMETRY KEY
1. What volume of dinitrogen oxide gas (nitrous oxide or laughing gas) at 18.0 °C and 786.3 mmHg will
form from the complete decomposition of 45.987 g ammonium nitrate?
NH4NO3(s)
→
N2O( g)
+
2H2O(l)
 1 mol NH4 NO3  1 mol N 2 O
? L N 2 O = 45.987 g NH4 NO3 

 80.0435 g NH4 NO3  1 mol NH4 NO3
  0.082058 L atm 


K mol


 291.15 K   760 mmHg 


 = 13.27 L N 2 O
 786.3 mmHg   1 atm 
2. A important reaction in the production of nitrogen fertilizers is the oxidation of ammonia to form
nitrogen monoxide and water at 500 °C.
a. How many liters of nitrogen monoxide will be formed at the same time as 26,500 L of water vapor
if both are measured at 500 °C and 762.5 mmHg?
4NH3 + 5O2
→
4NO + 6H2O
 4 L NO 
? L NO = 26,500 L H2 O 

 6 L H2 O 
= 17,667 L NO or 1.77 × 104 L NO if 26,5000 is three significant figures.
b. How many liters of oxygen at 25 °C and 0.956 atm are necessary to produce 385 L of NO(g) at
500 °C and 765 mmHg?
K mol

 765 mmHg   1 atm 
? L O2 = 385 L NO 



 0.082058 L atm  773 K   760 mmHg 
 5 mol O2   0.082058 L atm   298 K 

 = 195 L O 2


K mol
  0.956 atm 
 4 mol NO  
3. 34.6 L of carbon dioxide at 324.6 K and 792.4 mmHg form in the complete combustion of
naphthalene, C10H8. What volume of oxygen at 275.4 K and 754.9 mmHg must have been consumed
in this reaction?
C10H8 + 12O2
→
10CO2 + 4H2O
K mol

  792.4 mmHg   1 atm 
? L O2 = 34.6 L CO2 



 0.082058 L atm  324.6 K   760 mmHg 
 12 mol O2

 10 mol CO2
  0.082058 L atm   275.4 K   760 mmHg 



 = 37.0 L O 2
K mol
  754.9 mmHg   1 atm 

383
DALTON’S LAW OF PARTIAL PRESSURES WORKSHEET KEY
1. A sample of hydrogen gas is collected over water. The total pressure of the wet hydrogen in a 500 mL
container is 756.5 mmHg at 21 °C. If the hydrogen is dried and placed in a 200 mL container at 18 °C,
what will its pressure be?
Ptotal = PH2 + PH2O vap
initial PH2 = Ptotal − PH2Ovap = 756.5 mmHg − 18.7 mmHg = 737.8 mmHg
 V  T 
P2 = P1  1  2  = 737.8 mmHg
 V2  T1 
P1 V1
PV
= 2 2
T1
T2
 291 K   500 mL 
3
 = 1.83 x 10 mmHg


 294 K   200 mL 
To check: decreased V leads to increased P, so the volume ratio should be greater than one to lead to the
expected increase in pressure. Decreased T leads to decreased P, so the temperature ratio should be less
than one to lead to the expected decrease in pressure.
2. A 252 mL container has 0.082 g of hydrogen, 0.15 g of oxygen and 0.092 g of carbon dioxide. The
temperature is 22.52 °C. What is the total pressure of the gas in the container?
Ptotal =
each
∑P
gas
partial
 each  RT
RT
= ∑n
= (n H2 + nO2 + nCO2 )
V
 gas  V

 1 mol H2 
 1 mol O2 
 1 mol CO2  
= 0.082 g H2 
 + 0.15 g O2 
 + 0.092 g CO2 

 2.0159 g H2 
 31.9988 g O2 
 44.010 g CO2  

 0.082058 L atm 295.67 K 
(
)

K mol

 = 4.6 atm
0.252
L




3. A container holds a mixture of 0.0057 g of neon and 0.0081 g of an unknown gas. The partial pressure of
the neon is 500.4 mmHg and the total pressure is 755.2 mmHg. What is the molar mass of the
unknown gas?
PNe = X Ne PT =
n Ne
PT
n Ne + n unk
n Ne + n unk =
PT n Ne
PNe
n unk =
PT n Ne
− n Ne
PNe

 1 mol Ne  
755.2 mmHg 0.0057 g Ne 

 1 mol Ne 
 20.1797 g Ne  

−4
=
- 0.0057 g Ne 
 = 1.4 x 10 mol unk
500.4 mmHg
 20.1797 g Ne 
? g unk
0.0081 g unk
=
= 58 g/mol
mol unk 1.4 x 10−4 mol unk
384
Chapter 10 Worksheet Keys
CHEMISTRY 151 - DALTON’S LAW OF PARTIAL PRESSURES KEY
1. A 500.0 mL sample of a gas is collected over water at 30 ºC and 0.923 atm. What volume would the gas
occupy if dry and at 100 ºC and 1.000 atm?
 1 atm 
Pinitial = Ptotal − Pwater vapor = 0.923 atm − 31.824 mmHg 
 = 0.881 atm
 760 mmHg 
373 K   0.881 atm 
V2 = 500.0 mL 
 = 542 mL

 303 K   1.000 atm 
2. A cyclopropane-oxygen mixture is used as an anesthetic. Cyclopropane is C3H6. If 24.0 g of
cyclopropane and 48.0 g of oxygen are placed in a 5.0 L container at 18.5 ºC, what will the total pressure
of the gases be?
 each  RT
P = ∑n
 gas  V

 1 mol C3 H6 
 1 mol O2
= 24.0 g C3 H6 
 + 48.0 g O2 
 31.9988 g O2
 42.081 g C3 H6 

= 9.9 atm
L atm
0.082058
(291.7 K )


K mol

5.0 L
 
3. A mixture of 0.024 g of N2O(g ) and 0.825 g NO(g) exerts a pressure of 1.32 atm. What is the partial
pressure of each gas?
1 mol N 2 O
0.024 g N 2 O
44.0128 g N 2 O
X N2O =
= 0.019
1 mol N 2 O
1 mol NO
0.024 g N 2 O
+ 0.825 g NO
44.0128 g N 2 O
30.0061 g NO
PN2O = X N2O PT = 0.019 (1.32 atm ) = 0.025 atm N 2 O
PNO = 1.32 atm − 0.025 atm = 1.29 atm NO
4. The atmosphere in a sealed diving bell contained oxygen and helium. If the gas mixture has 0.200 atm of
oxygen and a total pressure of 3.00 atm, calculate the mass of helium in 1.00 L of the gas mixture at
20 ºC.
PHe = PT − PO2 = 3.00 atm − 0.200 atm = 2.80 atm He
PV =
g
RT
M
g=
PVM
=
RT
g
mol = 0.466 g He
L atm
0.082058
(293 K )
K mol
2.80 atm (1.00 L ) 4.0026
5. A quantity of nitrogen gas originally at 4.60 atm in a 1.20 L container at 23.9 ºC is transferred to a
10.0-L container. A quantity of oxygen originally at 3.50 atm and 23.9 ºC in a 4.00 L container is transferred to the same 10.0 L container with the nitrogen. What is the total pressure of the two gases in the
10.0 L container at 19.7 ºC?
Ptotal = PN2 + PO2 = 0.544 atm + 1.38 atm = 1.92 atm total
1.20 L   292.9 K 
PN2 = 4.60 atm 

 = 0.544 atm
 10.0 L   297.1 K 
4.00 L   292.9 K 
PO2 = 3.50 atm 

 = 1.38 atm
 10.0 L  297.1 K 