Combined Gas Laws EquationsThe Volume-Pressure (Boyle's) Law states that for a sample of gas at constant temperature, pressure is directly proportional to 1/V:P1/V or V1/P

The Volume-Temperature (Charles') Law states that for a sample of gas at constant pressure, volume is directly proportional to the absolute (Kelvin) temperature:VT

Combining these givesVT1/P or VT /P or V = cT/P where c is a proportionality constant.

Rearranging this gives PV/T = cSince PV/T is always constant, for a gas sample at an intial pressure, volume and temperature of P1, V1and T1and a final pressure, volume and temperature of P2, V2and T2this becomes:P1V1 = P2V2

T1T2

If five out of six of these quantities are known, it is possible to solve this equation for the sixth quantity.If the unknown quantity is V2, the equation becomes:V2 = V1P1T2

P2T1

If the unknown quantity is P2, the equation becomes:P2 = P1V1T2

V2T1

If the unknown quantity is T2, the equation becomes:T2 = T1P2V2

P1V1

A qualitative approach to gas law problems.When you are solving numerical problems using the gas laws, you can often check to be sure that the magnitude of your answer makes sense.If you are using the Volume-Temperature Law: V = a TAt constant pressure, volume is directly proportional to temperature. If the temperature increases the volume must also increase and your final volume should be larger than your initial volume. If this is not the case, you have made an error.Similarly, if the temperature decreases the volume must also decrease, and your final volume should be less than your initial volume.If you are using the Volume-Pressure Law: PV = bAt constant temperature, volume is inversely proportional to pressure. If the pressure increases the volume must decrease and your final volume should be less than your initial volume. If this is not the case, you have made an error.Similarly, if the pressure decreases the volume must increase and your final volume should be larger than your initial volume.If you are using the combined gas laws: PV/T = cIn this case it is not always possible to decide on relative sizes, because there are two variables that can work in opposite directions.If the temperature increases while the pressure decreases, then both changes cause the volume to increase and the final volume will be larger than the initial volume.If the temperature decreases while the pressure increases, then both changes cause the volume to decrease and the final volume will be smaller than the inital volume.If the temperature and pressure both change in the same direction, then it is not possible to predict what will happen without doing a calculation to see which change dominates.If you solve PV/T = c for pressure, you get P = cT/V which says that pressure is directly proportional to temperature and inversely proportional to volume. It is possible to use a similar type of reasoning to decide what should happen to the pressure when the temperature and volume are changed.

Since volume and pressure are inversely proportional (P1/V),increasingthe volume alone willdecreasethe pressure.

Since pressure is directly proportional to temperature (PT),decreasingthe temperature alone willdecreasethe pressure.

A sample of gas contains0.1100mol ofN2(g)and0.2200mol ofH2(g)and occupies a volume of12.4L. The following reaction takes place:

N2(g)+2H2(g)N2H4(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.4.13L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1100molN2(g)and the0.2200molH2(g)are present in amounts that react completely:

N2(g)+2H2(g)N2H4(g)0.1100molN2(g)2molH2(g) = 0.2200molH2(g)

1molN2(g)

Similarly, the balanced equation shows the amount of product that forms:0.1100molN2(g)1molN2H4(g) = 0.1100molN2H4(g)

1molN2(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1100mol +0.2200mol =0.3300mol
V1=12.4L
n2=0.1100mol
V2= ?V2= V1n2 = (12.4L)(0.1100mol) =4.13L

n10.3300mol

A sample of gas contains0.1700mol ofHCl(g)and8.500E-2mol ofBr2(g)and occupies a volume of11.5L. The following reaction takes place:

2HCl(g)+Br2(g)2HBr(g)+Cl2(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.11.5L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1700molHCl(g)and the0.0850molBr2(g)are present in amounts that react completely:

2HCl(g)+Br2(g)2HBr(g)+Cl2(g)0.1700molHCl(g)1molBr2(g) = 0.0850molBr2(g)

2molHCl(g)

Similarly, the balanced equation shows the amounts of products that form:0.1700molHCl(g)2molHBr(g) = 0.1700molHBr(g)

2molHCl(g)

0.1700molHCl(g)1molCl2(g) = 0.0850molCl2(g)

2molHCl(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1700mol +0.0850mol =0.2550mol
V1=11.5L
n2=0.1700mol +0.0850mol =0.2550mol
V2= ?V2= V1n2 = (11.5L)(0.2550mol) =11.5L

n10.2550mol

A sample of gas contains0.1100mol ofN2(g)and0.2200mol ofO2(g)and occupies a volume of11.0L. The following reaction takes place:

N2(g)+2O2(g)2NO2(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.7.33L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1100molN2(g)and the0.2200molO2(g)are present in amounts that react completely:

N2(g)+2O2(g)2NO2(g)0.1100molN2(g)2molO2(g) = 0.2200molO2(g)

1molN2(g)

Similarly, the balanced equation shows the amount of product that forms:0.1100molN2(g)2molNO2(g) = 0.2200molNO2(g)

1molN2(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1100mol +0.2200mol =0.3300mol
V1=11.0L
n2=0.2200mol
V2= ?V2= V1n2 = (11.0L)(0.2200mol) =7.33L

n10.3300mol

A sample of gas contains0.1600mol ofN2(g)and0.3200mol ofH2(g)and occupies a volume of21.7L. The following reaction takes place:

N2(g)+2H2(g)N2H4(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.7.23L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1600molN2(g)and the0.3200molH2(g)are present in amounts that react completely:

N2(g)+2H2(g)N2H4(g)0.1600molN2(g)2molH2(g) = 0.3200molH2(g)

1molN2(g)

Similarly, the balanced equation shows the amount of product that forms:0.1600molN2(g)1molN2H4(g) = 0.1600molN2H4(g)

1molN2(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1600mol +0.3200mol =0.4800mol
V1=21.7L
n2=0.1600mol
V2= ?V2= V1n2 = (21.7L)(0.1600mol) =7.23L

n10.4800mol

A sample of gas contains0.1200mol ofCO(g)and0.1200mol ofCl2(g)and occupies a volume of8.38L. The following reaction takes place:

CO(g)+Cl2(g)COCl2(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.4.19L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1200molCO(g)and the0.1200molCl2(g)are present in amounts that react completely:

CO(g)+Cl2(g)COCl2(g)0.1200molCO(g)1molCl2(g) = 0.1200molCl2(g)

1molCO(g)

Similarly, the balanced equation shows the amount of product that forms:0.1200molCO(g)1molCOCl2(g) = 0.1200molCOCl2(g)

1molCO(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1200mol +0.1200mol =0.2400mol
V1=8.38L
n2=0.1200mol
V2= ?V2= V1n2 = (8.38L)(0.1200mol) =4.19L

n10.2400mol

A sample of gas contains0.1900mol ofCH4(g)and0.1900mol ofH2O(g)and occupies a volume of14.6L. The following reaction takes place:

CH4(g)+H2O(g)3H2(g)+CO(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.29.2L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1900molCH4(g)and the0.1900molH2O(g)are present in amounts that react completely:

CH4(g)+H2O(g)3H2(g)+CO(g)0.1900molCH4(g)1molH2O(g) = 0.1900molH2O(g)

1molCH4(g)

Similarly, the balanced equation shows the amounts of products that form:0.1900molCH4(g)3molH2(g) = 0.5700molH2(g)

1molCH4(g)

0.1900molCH4(g)1molCO(g) = 0.1900molCO(g)

1molCH4(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1900mol +0.1900mol =0.3800mol
V1=14.6L
n2=0.5700mol +0.1900mol =0.7600mol
V2= ?V2= V1n2 = (14.6L)(0.7600mol) =29.2L

n10.3800mol

A sample of gas contains0.1500mol ofN2(g)and0.3000mol ofO2(g)and occupies a volume of22.7L. The following reaction takes place:

N2(g)+2O2(g)2NO2(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.15.1L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1500molN2(g)and the0.3000molO2(g)are present in amounts that react completely:

N2(g)+2O2(g)2NO2(g)0.1500molN2(g)2molO2(g) = 0.3000molO2(g)

1molN2(g)

Similarly, the balanced equation shows the amount of product that forms:0.1500molN2(g)2molNO2(g) = 0.3000molNO2(g)

1molN2(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1500mol +0.3000mol =0.4500mol
V1=22.7L
n2=0.3000mol
V2= ?V2= V1n2 = (22.7L)(0.3000mol) =15.1L

n10.4500mol

A sample of gas contains0.1500mol ofCO(g)and0.1500mol ofCl2(g)and occupies a volume of10.9L. The following reaction takes place:

CO(g)+Cl2(g)COCl2(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.5.45L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1500molCO(g)and the0.1500molCl2(g)are present in amounts that react completely:

CO(g)+Cl2(g)COCl2(g)0.1500molCO(g)1molCl2(g) = 0.1500molCl2(g)

1molCO(g)

Similarly, the balanced equation shows the amount of product that forms:0.1500molCO(g)1molCOCl2(g) = 0.1500molCOCl2(g)

1molCO(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1500mol +0.1500mol =0.3000mol
V1=10.9L
n2=0.1500mol
V2= ?V2= V1n2 = (10.9L)(0.1500mol) =5.45L

n10.3000mol

A sample of gas contains0.1900mol ofCH4(g)and0.3800mol ofO2(g)and occupies a volume of19.7L. The following reaction takes place:

CH4(g)+2O2(g)CO2(g)+2H2O(g)

Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.19.7L

Feedback:
Avogadro's Law says that equal volumes of gases at the same temperature and pressure contain the same number of molecules (or moles). Mathematically this means that at a given temperature and pressure, the volume of a gas sample is directly proportional to the number of moles of gas in the sample:

V = constantn

orV1 = V2 = constant

n1n2

From the stoichiometry of the balanced equation, it can be seen that the0.1900molCH4(g)and the0.3800molO2(g)are present in amounts that react completely:

CH4(g)+2O2(g)CO2(g)+2H2O(g)0.1900molCH4(g)2molO2(g) = 0.3800molO2(g)

1molCH4(g)

Similarly, the balanced equation shows the amounts of products that form:0.1900molCH4(g)1molCO2(g) = 0.1900molCO2(g)

1molCH4(g)

0.1900molCH4(g)2molH2O(g) = 0.3800molH2O(g)

1molCH4(g)

The intial and final volumes are directly proportional to the numbers of moles:

n1=0.1900mol +0.3800mol =0.5700mol
V1=19.7L
n2=0.1900mol +0.3800mol =0.5700mol
V2= ?V2= V1n2 = (19.7L)(0.5700mol) =19.7L

n10.5700mol

A sample ofhydrogengas has a density of3.81E-2g/L at a pressure of0.512atm and a temperature of58oC.Assume ideal behavior.

Feedback:

From the ideal gas law PV=nRT, the density of a gas in moles per liter is:n = P

VRT

where P is the pressure in atm, R = 0.08206 L atm/mol K and T is in Kelvins.

In this case:
P =0.512atm
T =58oC + 273 =331Kn = 0.512atm = 1.88E-2mol/L

V0.08206 L atm/mol K331K

The molecular weight ofhydrogenis2.02g/mol. Multiplying mol/L by g/mol gives g/L.1.88E-2mol2.02g= 3.81E-2g

LmolL

A sample of an unknown gas is found to have a density of1.46g/L at a pressure of0.546atm and a temperature of49oC. The molecular weight of the unknown gas is70.7g/mol.

Assume ideal behavior.

Feedback:

Method 1:Assume a volume of 1 L, use the ideal gas law to solve for the number of moles, and take the ratio g/L to mol/L to get the molar mass:n = PV = (0.546atm)(1 L) = 2.066E-2mol in 1 L

RT(0.08206L atm mol-1K-1)(322K)

Molecular weight = 1.46g/L = 70.7g/mol

2.066E-2mol/L

Method 2:Use a formula.

The number of moles of gas can be calculated as g/M, where g is the mass in grams and M is the molar mass in g/mol. Substituting g/M for n into the ideal gas law PV=nRT gives PV =(g/M)RT which can be rearranged to give g/V = PM/RT. Since g/V is the density, d, in grams/L , the ideal gas law in terms of density and molar mass becomes:d = PM

RT

where P is the pressure in atm, R = 0.08206 L atm/mol K, and T is in Kelvins. Rearranging to solve for the molar mass of the gas givesM = dRT

P

Substituting the values
d =1.46g/L
R = 0.08206 L atm/mol K
T =49oC + 273 =322K
P =0.546atm

givesM = 1.46g0.08206 L atm322K 1 = 70.7g/mol

Lmol K0.546atm

What volume ofhydrogen sulfideis required to produce86.2liters ofsulfur dioxideaccording to the following reaction? (All gases are at the same temperature and pressure.)

hydrogen sulfide(g)+oxygen(g)water(l)+sulfur dioxide(g)

86.2litershydrogen sulfide

Feedback:
1.Write a balanced chemical equation for the above reaction:
2H2S(g)+3O2(g)2H2O(l)+2SO2(g)


2.The number of moles of a gas is proportional to the number of liters of the gas. For two gases at the same temperature and pressure, the volume ratio is the same as the mole ratio.


3.Determine the number of liters ofhydrogen sulfiderequired:LH2S= 86.2LSO22LH2S = 86.2LH2S

2LSO2

What volume ofhydrogen gasis produced when2.16mol ofsodiumreacts completely according to the following reaction at 0oC and 1 atm?

sodium(s)+water(l)sodium hydroxide(aq)+hydrogen(g)

24.2litershydrogen gas

Feedback:
1.Write a balanced chemical equation for the above reaction:
2Na(s)+2H2O(l)2NaOH(aq)+H2(g)


2.Determine the number of moles ofhydrogen gasproduced:molH2= 2.16molNa1molH2 = 1.08molH2

2molNa

3.Determine the number of liters ofhydrogen gasproduced:
The volume of one mole of gas at 0oC and 1 atm is 22.4 L.LH2= 1.08molH222.4 LH2 = 24.2LH2

molH2

How many moles ofhydrogen peroxide (H2O2)are needed to produce18.0L ofoxygen gasaccording to the following reaction at 0oC and 1 atm?

hydrogen peroxide (H2O2)(aq)water(l)+oxygen(g)

1.61moleshydrogen peroxide (H2O2)

Feedback:
1.Write a balanced chemical equation for the above reaction:
2H2O2(aq)2H2O(l)+O2(g)


2.Determine the number of moles ofoxygen gasproduced:
The volume of one mole of gas at 0oC and 1 atm is 22.4 L.molO2= 18.0LO21 molO2 = 0.804molO2

22.4 LO2

3.Determine the number of moles ofhydrogen peroxide (H2O2)needed:molH2O2= 0.804molO22molH2O2 = 1.61molH2O2

1molO2

Nitrogen monoxide is produced by combustion in an automobile engine.How many moles ofnitrogen monoxideare required to react completely with17.1L ofoxygen gasaccording to the following reaction at 0oC and 1 atm?

nitrogen monoxide(g)+oxygen(g)nitrogen dioxide(g)

1.52molesnitrogen monoxide

Feedback:
1.Write a balanced chemical equation for the above reaction:
2NO(g)+O2(g)2NO2(g)


2.Determine the number of moles ofoxygen gasavailable:
The volume of one mole of gas at 0oC and 1 atm is 22.4 L.molO2= 17.1LO21 molO2 = 0.761molO2

22.4 LO2

3.Determine the number of moles ofnitrogen monoxiderequired:molNO= 0.761molO22molNO = 1.52molNO

1molO2

What volume ofhydrogen gasis produced when2.25mol ofsodiumreacts completely according to the following reaction at 0oC and 1 atm?

sodium(s)+water(l)sodium hydroxide(aq)+hydrogen(g)

25.2litershydrogen gas

Feedback:
1.Write a balanced chemical equation for the above reaction:
2Na(s)+2H2O(l)2NaOH(aq)+H2(g)


2.Determine the number of moles ofhydrogen gasproduced:molH2= 2.25molNa1molH2 = 1.13molH2

2molNa

3.Determine the number of liters ofhydrogen gasproduced:
The volume of one mole of gas at 0oC and 1 atm is 22.4 L.LH2= 1.13molH222.4 LH2 = 25.2LH2

molH2

How many grams ofcalcium carbonateare needed to produce22.6L ofcarbon dioxideaccording to the following reaction at 25oC and 1 atm?

calcium carbonate(s)calcium oxide(s)+carbon dioxide(g)

92.3gramscalcium carbonate

Feedback:
1.Write a balanced chemical equation for the above reaction:
CaCO3(s)CaO(s)+CO2(g)


2.Determine the number of moles ofcarbon dioxideproduced:
From PV=nRT, n=PV/RTmolCO2= 1 atm22.6Lmol K1 = 0.922molCO2

0.08206 L atm298K

3.Determine the number of moles ofcalcium carbonateneeded:molCaCO3= 0.922molCO21molCaCO3 = 0.922molCaCO3

1molCO2

4.Determine the number of grams ofcalcium carbonateneeded:gCaCO3= 0.922molCaCO3100gCaCO3 = 92.3gCaCO3

molCaCO3

What volume ofhydrogen gasis produced when6.69g ofironreacts completely according to the following reaction at 25oC and 1 atm?

iron(s)+hydrochloric acid(aq)iron(II) chloride(aq)+hydrogen(g)

2.93litershydrogen gas

Feedback:
1.Write a balanced chemical equation for the above reaction:
Fe(s)+2HCl(aq)FeCl2(aq)+H2(g)


2.Determine the number of moles ofironavailable:molFe= 6.69gFe1 molFe = 0.120molFe

55.9gFe

3.Determine the number of moles ofhydrogen gasproduced:molH2= 0.120molFe1molH2 = 0.120molH2

1molFe

4.Determine the number of liters ofhydrogen gasproduced:
From PV=nRT, V=nRT/PLH2= 0.120mol0.08206 L atm298K1 = 2.93LH2

mol K1 atm

What volume ofcarbon dioxideis produced when70.8g ofcalcium carbonatereacts completely according to the following reaction at 25oC and 1 atm?

calcium carbonate(s)calcium oxide(s)+carbon dioxide(g)

17.3literscarbon dioxide

Feedback:
1.Write a balanced chemical equation for the above reaction:
CaCO3(s)CaO(s)+CO2(g)


2.Determine the number of moles ofcalcium carbonateavailable:molCaCO3= 70.8gCaCO31 molCaCO3 = 0.707molCaCO3

100gCaCO3

3.Determine the number of moles ofcarbon dioxideproduced:molCO2= 0.707molCaCO31molCO2 = 0.707molCO2

1molCaCO3

4.Determine the number of liters ofcarbon dioxideproduced:
From PV=nRT, V=nRT/PLCO2= 0.707mol0.08206 L atm298K1 = 17.3LCO2

mol K1 atm

How many grams ofsodiumare needed to produce46.3L ofhydrogen gasaccording to the following reaction at 25oC and 1 atm?

sodium(s)+water(l)sodium hydroxide(aq)+hydrogen(g)

87.0gramssodium

Feedback:
1.Write a balanced chemical equation for the above reaction:
2Na(s)+2H2O(l)2NaOH(aq)+H2(g)


2.Determine the number of moles ofhydrogen gasproduced:
From PV=nRT, n=PV/RTmolH2= 1 atm46.3Lmol K1 = 1.89molH2

0.08206 L atm298K

3.Determine the number of moles ofsodiumneeded:molNa= 1.89molH22molNa = 3.78molNa

1molH2

4.Determine the number of grams ofsodiumneeded:gNa= 3.78molNa23.0gNa = 87.0gNa

molNa

The stopcock connecting a1.55L bulb containingxenongas at a pressure of8.34atm, and a2.24L bulb containingneongas at a pressure of4.59atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is6.12atm.

Feedback:
The total pressure in the final mixture is given by the sum of the partial pressures of the two gases:

Ptotal= PXe+ PNe

To find the partial pressure of each gas, treat each as if it were alone and expanded into the final volume.

Then: P1V1= P2V2or P2= P1(V1/V2) where V2= the combined volume of both bulbs.

Substituting in for each gas:P2(Xe) = 8.34atm 1.55L = 3.41atm

(1.55+2.24) L

P2(Ne) = 4.59atm 2.24L = 2.71atm

(1.55+2.24) L

A mixture ofmethaneandneongases, in a7.48L flask at54oC, contains4.52grams ofmethaneand3.36grams ofneon. The partial pressure ofneonin the flask is0.597atm and the total pressure in the flask is1.61atm.

Feedback:
The partial pressure of each gas can be calculated using the ideal gas law PAV = nART or PA= nART/V
where T, the Kelvin temperature =54oC + 273 =327K.

The number of moles of each gas must first be calculated from the number of grams and the molar mass:4.52gCH41 mol = 0.2825molCH4

16.0g

3.36gNe1 mol = 0.1663molNe

20.2g

The partial pressures can then be calculated:PCH4= nCH4RT = 0.2825mol 327K 0.08206 L atm1 = 1.01atm

Vmol K7.48L

PNe= nNeRT = 0.1663mol 327K 0.08206 L atm1 = 0.597atm

Vmol K7.48L

The total pressure is then the sum of the partial pressures:

Ptotal=1.01atm +0.597atm =1.61atm

Alternatively, the total pressure can be calculated using the ideal gas law and the total number of moles.

ntot= nCH4+ nNe=0.2825mol +0.1663mol =0.4488molPtot= ntotRT = 0.4488mol 327K 0.08206 L atm1 = 1.61atm

Vmol K7.48L

A mixture ofxenonandcarbon dioxidegases is maintained in a6.38L flask at a pressure of2.39atm and a temperature of46oC. If the gas mixture contains29.0grams ofxenon, the number of grams ofcarbon dioxidein the mixture is15.9g.

Feedback:
Since the total pressure, the temperature, and the volume are all known, the total number of moles of gas in the mixture can be calculated using the ideal gas law:

PtotalV = ntotalRT
where T, the Kelvin temperature =46oC + 273 =319K and ntotal= nXe+ nCO2.

Thenntotal= PtotalV = 2.39atm 6.38L mol K1 = 0.5825mol

RT0.08206 L atm319K

The number of moles ofXecan be calculated from the number of grams ofXeand the molar mass:29.0gXe1 mol = 0.2214molXe

131g

The number of moles ofCO2can then be calculated from the total number of moles:

nCO2= ntotal- nXe=0.5825-0.2214=0.3611molCO2

And finally, the number of grams ofCO2can be calculated:0.3611molCO244.0g = 15.9gCO2

mol

A mixture ofheliumandmethanegases is maintained in a7.68L flask at a pressure of1.34atm and a temperature of22oC. If the gas mixture contains0.726grams ofhelium, the number of grams ofmethanein the mixture is3.90g.

Feedback:
Since the total pressure, the temperature, and the volume are all known, the total number of moles of gas in the mixture can be calculated using the ideal gas law:

PtotalV = ntotalRT
where T, the Kelvin temperature =22oC + 273 =295K and ntotal= nHe+ nCH4.

Thenntotal= PtotalV = 1.34atm 7.68L mol K1 = 0.4251mol

RT0.08206 L atm295K

The number of moles ofHecan be calculated from the number of grams ofHeand the molar mass:0.726gHe1 mol = 0.1815molHe

4.00g

The number of moles ofCH4can then be calculated from the total number of moles:

nCH4= ntotal- nHe=0.4251-0.1815=0.2436molCH4

And finally, the number of grams ofCH4can be calculated:0.2436molCH416.0g = 3.90gCH4

mol

A mixture ofheliumandnitrogengases, in a9.36L flask at46oC, contains0.869grams ofheliumand2.75grams ofnitrogen. The partial pressure ofnitrogenin the flask is0.275atm and the total pressure in the flask is0.882atm.

Feedback:
The partial pressure of each gas can be calculated using the ideal gas law PAV = nART or PA= nART/V
where T, the Kelvin temperature =46oC + 273 =319K.

The number of moles of each gas must first be calculated from the number of grams and the molar mass:0.869gHe1 mol = 0.2173molHe

4.00g

2.75gN21 mol = 9.821E-2molN2

28.0g

The partial pressures can then be calculated:PHe= nHeRT = 0.2173mol 319K 0.08206 L atm1 = 0.608atm

Vmol K9.36L

PN2= nN2RT = 9.821E-2mol 319K 0.08206 L atm1 = 0.275atm

Vmol K9.36L

The total pressure is then the sum of the partial pressures:

Ptotal=0.608atm +0.275atm =0.882atm

Alternatively, the total pressure can be calculated using the ideal gas law and the total number of moles.

ntot= nHe+ nN2=0.2173mol +9.821E-2mol =0.3155molPtot= ntotRT = 0.3155mol 319K 0.08206 L atm1 = 0.882atm

Vmol K9.36L

A1.54mol sample ofO2gas is confined in a38.2liter container at28.9oC.

If the volume of the gas sample isdecreasedto19.1L, holding the temperature constant, the pressure willincrease. Which of the following kinetic theory ideas apply?

Choose all that apply.With less available volume, the molecules hit the walls of the container more often.

For a given gas at constant temperature, the force per collision is constant. Some other factor must cause the pressure increase.

With higher average speeds, on average the molecules hit the walls of the container with more force.

At lower volumes molecules have higher average speeds.

None of the Above

Feedback:
Decreasing the volume at constant temperature increases the gas pressure because the number of molecule-wall collisions per unit area per unit time increases. For a given gas, both the average molecular speed and the average force per collision are constant at constant temperature.

A1.91mol sample ofXegas is confined in a48.6liter container at36.8oC.

If the volume of the gas sample is decreased to24.3L holding the temperature constant, thenumber of molecule-wall collisions per unit area per unit timewillincrease

decrease

remain the same

not enough information to answer the question

Feedback:
Since the temperature is constant, both the average kinetic energy and the average speed of the molecules is constant. However, if the volume is decreased the molecules do not have as far to travel to collide with the walls, so the number of molecule-wall collisions will increase.

A1.91mol sample ofKrgas is confined in a46.2liter container at21.8oC.

If the volume of the gas sample is increased to92.4L holding the temperature constant, theaverage force per molecule-wall collisionwilldecrease

remain the same

not enough information to answer the question

increase

Feedback:
Since the temperature is constant, the average speed of the molecules is constant. Although there will be fewer molecule-wall collisions due to the increased volume, the average force per collision will be the same since the speed and therefore the momentum are constant.

A1.04mol sample ofXegas is confined in a25.3liter container at23.0oC.

If the volume of the gas sample isdecreasedto12.6L, holding the temperature constant, the pressure willincrease. Which of the following kinetic theory ideas apply?

Choose all that apply.With higher average speeds, on average the molecules hit the walls of the container with more force.

At lower volumes molecules have higher average speeds.

For a given gas at constant temperature, only the number of molecule-wall collisions has an effect on pressure.

With less available volume, the molecules hit the walls of the container more often.

None of the Above

Feedback:
Decreasing the volume at constant temperature increases the gas pressure because the number of molecule-wall collisions per unit area per unit time increases. For a given gas, both the average molecular speed and the average force per collision are constant at constant temperature.
A1.50mol sample ofN2gas is confined in a37.4liter container at31.2oC.

If the volume of the gas sample is decreased to18.7L holding the temperature constant, theaverage molecular speedwilldecrease

remain the same

not enough information to answer the question

increase

Feedback:
Since the temperature is constant, both the average kinetic energy and the average speed of the molecules is constant.

A1.25mol sample ofN2gas is confined in a30.7liter container at26.0oC.

If the volume of the gas sample is increased to61.3L holding the temperature constant, theaverage force per molecule-wall collisionwilldecrease

increase

not enough information to answer the question

remain the same

Feedback:
Since the temperature is constant, the average speed of the molecules is constant. Although there will be fewer molecule-wall collisions due to the increased volume, the average force per collision will be the same since the speed and therefore the momentum are constant.

Arrange the following molecules in order of increasing average molecular speed. All are at the same temperature.

methane,nitrogen dioxide,krypton,neon

Enterformulasin the boxes below: 1 = slowest, 4 = fastest1234Correct answer:Kr

Kr

Your response:NO2

Correct answer:NO2

NO_2_

Your response:CH4

Correct answer:Ne

Ne

Your response:Kr

Correct answer:CH4

CH_4_

Your response:Ne

Feedback:
At a given temperature, the average molecular speed decreases with increasing molar mass. In this case:

CH4, MW =16.0
Ne, MW =20.2
NO2, MW =46.0
Kr, MW =83.8

Arranging these in order of decreasing molar mass, puts them in order of increasing average molecular speed.

Kr