Chapter 5: Ideal Gas and Ideal Gas Laws
The gaseous state is one of the states of matter. Gases have many properties. Gases can
occupy any shape, have high entropy(disorder) values and are very compressible.
Mixtures of gases are called solutions because gaseous mixtures are homogeneous. The
air we breath is a solution of-among other things- Nitrogen , oxygen and carbon dioxide
gases.
One important property of gases is pressure. Pressure is the force per unit area (Newton
per meter squared ). Pressure measurements can be in units of pascal(pa), torr, mmHg,
atm(atmosphere).
1.01325 x 105 pa = 1atm = 760 torr = 760 mmHg.
Open- End Manometer Pgas = 760 - 12 = 748mmHg
Gas Laws (gas particles as points in space w/elastic collisions)
(change the variable in red)
Boyle’s Law- V  1/P (constant n & T)
P1V1= k = P2V2
●→
●
Charles’s Law- V  T (constant n & P)
V = kT
Avogadro’s Law- V  n (constant P & T)
V = kn
V  (nT)/P
V = R(nT)/P
●→●
PV = nRT Ideal Gas Law
R is gas constant = 0.0821 (L-atm)/mol-K
OR
8.314 J/mol-K
V = Volume is always in liters.
T = Temperature is always in Kelvin
P = Pressure is usually in atmospheres (atm) ; (1 atm = 760 torr OR mmHg).
Standard Pressure & Temperature (STP):
1 atmosphere for pressure (1 atm)
273 K (0° C ) For temperature
Given 3 variables in the ideal gas equation, the 4th value can be calculated.
Ex. If a gas occupies V = 250ml, P = 1.30 atm, T = 31C, n = ?
n = (PV)/RT
n=
(1.3atm)(0.25L)
(0.0821 L-atm/mol-K)(304K)
n = 0.013 mole.
At standard temperature and pressure (STP, 1 atm, 0 C), 1 mole of every gas occupies a
volume of 22.4 liters (molar volume).
The ideal gas equation can be used to derive the gas laws of Boyle and Charles at two
different conditions and make calculation.
P1V1 = n1RT1
P2V2 = n2RT2
If n is constant, then
P1V1 = T1
P2V2 T2
What if n & T are constant?, (n & P, n & V)?
Using the ideal gas equation, one can relate the gas density(g/L) to its molar
mass(M).
PV = (g/M)RT
g/V = PM/RT = density = d
Ex. What is density of carbon tetrachloride gas at 714 torr and 125C?
M = 154g/mol
Change 714 torr to atm(714 torr/760 torr = 0.939 atm)
Change 125C to K(273 + 125 = 398 K)
d = (0.939atm x 154g)/(0.0821 x 398 K) = 4.43 g/L
Partial Pressures of Gases in a Mixture
The total pressure of a mixture of gases is equal to the sum of the partials pressures.
Pt = P1 + P2 + P3 ….
Pt = ( n1 + n2 + n3 ….)RT/V
Use mole fractions ( X1 = n1/ nt) to calculate partial pressures(P1).
P1 = X1 Pt
Ex. A mixture of gases has 2 moles O2, 7 moles N2, 0.4 mole H2 in 4 liter container.
What is the total pressure of the mixture and what is the partial pressure of H2 in mmHg?
T = 273 K
Amount of n = 2+7+0.4 = 9.4moles
Pt = (9.4moles)(0.0821)(273K)/4liters = 52.7atm = 52.7 x 760mmHg = 4 x 104mmHg
P(H2) = (0.4/9.4)(4 x 104 mmHg) = 1702 mmHg
Sometimes in an experiment at a certain temperature, a gas can be collected over a body
of water. The total pressure over the water is due to the gas and the water vapor from the
body of water. The partial pressure from the water vapor at that temperature can be
looked up in a chemistry textbook.
Root Mean Square Speed of a gas (u)- This is close in value to the average speed(
OR rate) of the gaseous molecules.
_______
u = (3RT/M); u is in m/s; M is molecular mass in kilograms; In this case R =
8.314J/mol-K
Graham’s Law of Effusion
r1/ r2 = ( M2/ M1)
Ex. What is molecular mass of a homonuclear diatomic gas that effuses at 0.355 times the
rate of O2?
rx/ r2 = 0.355 = ( M2/ Mx) = (32g/ Mx)
0.126 = 32g/ Mx
Mx = 32g/0.126 = 254g/mol  I2
Non-Ideal Gas Equation: (Pexp + a(n/V)2 )(V – nb) = nRT
Where a & b are “fudge” factors.
Some Ideal Gas problems
1. Perform the following conversions:
a) 0.357 atm to torr
b) 6.60x10-2 torr to atm
torr to pa d) 802 mmHg to atm
c) 745
1. a.) 0.357 atm│760 torr = 271.3 torr ; You Do Rest!
│1 atm
2. A barometer reads the atmospheric pressure 764.7 torr. A
sample of gas is placed in vessel attached to an open-end mercury
manometer. The level of mercury in the open-end arm of the
monometer has a height of 136.4 mm, and that in the arm that is in
contact with the gas has a height 103.8 mm. What is the pressure
of gas in atmosphere?
2. 136.4mm – 103.8mm = 32.6mm; Pgas = 764.7 + 32.6 = 797.3 mmHg
3. Write an equation or proportionality expression that expresses
each of the following statements:
a) Law of combining volumes explained by Avogadro,s hypothesis.
b) For a given quantity of gas, the product of pressure and
volume is proportional to absolute temperature.
c) For a given quantity of gas at constant temperature, the
product of pressure and volume is constant.
3. a.) V = kn
b.) PV = kT
c.) PV = k
4. Calculate the following quantities for an ideal gas:
a) the pressure , in atms, if 8.25x10-2 mol occupies 174 ml at -15
0
C;
b) the number of moles in 2.50 L at 37 0C and 725 torr.
c) the volume of a gas occupied by 6.72x10-3 mol at 145 0C and a
pressure of 59.0 torr;
d) the absolute temperature of the gas at which 9.87x10-2 mol
occupies 164 ml at 682 torr;
4. a) P =?
P = nRT = (8.25 x 10-2)(0.0821)(258K) = 10 atm.
V
0.174L
b.) n = PV =
725 torr x 2.5 L
= 0.094 mole
RT
760 torr x 0.0821 x 310K
c.) V = nRT =
P
6.72 x 10-3mol x 0.0821 x 418K x 760 torr = 2.97L
59 torr
d.) T = PV =
682 torr x 0.164 L
= 18 K
-2
nR
760 torr x 9.87 x10 x 0.0821
5. Consider 5.62 L chlorine gas, Cl2 ,at 740 torr and 33 0C.
a) What volume will the gas occupy at 107 0C and 680 torr?
b) What volume will the gas occupy at STP?
c) At what pressure will the volume be 5.00 L if the temperature is 67 0C?
d) At what pressure will the volume be twice as much as original, if
temperature remains constant?
e) How many moles Cl2 gas are in original condition?
5. a.)
b.)
P1V1 = P2V2 = 5.62 L x740 torr = V2 x 680 torr ; V2 = 7.57 L
T1
T2
306 K
380 K
P1V1 = P2V2 = 5.62 L x740 torr = V2 x 760 ; V2 = 4.88 L
T1
T2
306 K
273 K
c.)
P1V1 = P2V2 = 5.62 L x740 torr = 5.0 L x P2 ; P2 = 924 torr
T1
T2
306 K
340 K
e.) YOU DO REST!
6. Perform the following calculations;
a) the density of SO3(g) at 0.96 atm and 35 0C.
b) the density of SO3(g) at STP condition.
c) the molar mass of a gas if its density is 3.67 g/L at 15 0C and 825 torr.
d) the molar mass of gas if 4.40 g occupies 3.50 L at 560 torr and 41 0C.
e) the molar mass of a gas if its density is 2.86 g/L at STP and identify the
molecular formula of the gas.
6. a.) Molecular wt. SO3 is 80g; D = PM = 0.96 atm x 80g = 3.04 g/L
RT
0.0821 x 308 K
b.)
D = PM = 1.00 atm x 80g = 3.57 g/L
RT
0.0821 x 273 K
c.) M =
DRT
P
YOU DO REST!(answer = 80 g/mol)
7. Consider the following reaction of ammonia gas and oxygen gas; at 850 0C
and 5.00 atm.
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
a) How many liters NH3(g) at 850 0C and 5.00 atm are required to react with
1.00 mol of O2(g) in this reaction?
b) How many moles NO(g) can be prepared using 450 L of NH3(g) at a
pressure of 5.00 atm and a temperature of 295 K?
c) How many grams of NH3(g) are needed to generate 10.0 L of NO(g) if
pressure of NO gas is 740 torr at 23 0C?
7. a.) Stoichiometricaly, every 1.00 mole O2 reacted => 0.8 mole NH3 reacted.
V=
nRT = 0.8 mol x 0.0821 x 1123 K = 14.75 L
P
5 atm
b. )Stoichiometrically, every 1.00 mole NH3 reacted => 1.00 mole NO formed.
Moles NH3 = PV = 5 atm x 450 L = 92.9 moles NO
RT
0.0821 x 295 K
c.) Stoichiometrically, every 1.00 mole NH3 reacted => 1.00 mole NO formed.
Moles NO = PV = 0.97 atm x 10 L = 0.4 moles NO => moles NH3
RT
0.0821x 296 K
0.4mol x 17g/mol = 6.8g NH3
8. Perform the following calculations;
a) the partial pressure and total pressure of each of the following gases in
the mixture of 4.00 g of each gas CH4(g), C2H4(g) and C4H10(g), in 1.50 L
flask at 0 0C.
b) the partial pressure of each component in a mixture of gases contains
0.55 mol N2, 0.20 mol O2, and 0.10 mol CO2. The total pressure of the
mixture is 1.32 atm.
c) the partial pressure of each component in a mixture of gases contains
3.50 g N2, 2.15 g H2, and 5.27 g of NH3. If the total pressure of the
mixture is 2.50 atm.
8. a.) 4g of each gas means :
0.25mole CH4, 0.143mole C2H4, 0.069mole C4H10.
Total mole = 0.25 + 0.143 + 0.069 = 0.462 mole
Total volume (Vtotal) is 1.5 L
Mole fraction (X) = mole of gas divided by total mole.
Partial pressure = (Xgas )(Ptotal)
Total pressure (Ptotal) = nRT 0.462 x 0.0821 x 273 K = 6.9atm
Vtotal
1.5 L
Partial pressure of C2H4 ( PC2H4) = 0.143 x 6.9 atm = 2.14 atm
0.462
Partial pressure of CH4 ( PCH4) = 0.25 x 6.9 atm = 3.73 atm
0.462
Partial pressure of C4H10 ( PC4H10) = 0.069 x 6.9 atm = 1.03 atm
0.462
b.) Total moles = 0.55 + 0.2 + 0.1 = 0.85 mole, Ptotal is 1.32 atm
Partial pressure of N2 ( PN2) =
0.55 x 1.32atm = 0.854 atm
0.85
Partial pressure of O2 ( PO2) = 0.2 x 1.32atm = 0.31atm
0.85
Partial pressure of CO2 ( PCO2) = 0.1 x 1.32 atm = 0.155atm
0.85
YOU DO part c.!
9. Consider the following gases at 25 0C; CO2, He, Cl2, CO, CH4, N2
a) place them in order of increasing average molecular speed.
b) calculate and compare the rms speeds of CO(g) and CO2(g).
c) calculate the ratio of rate of diffusion of gases CH4(g) and He.
d) which gas effuses 1.41 times faster than O2(g)?
e) which two gases have the same rate of diffusion?
9. The larger the mass, the slower the gas! Increasing molecular speed:
a.) Cl2 < CO2 < CO , N2 < CH4 < He
________
b.) µ =
√
√(3RT)/M ; M is in kilograms, R is 8.314 J/mol-K
means square root
________________________
For CO µ =
√(3 x 8.314 x 298 K)/ 0.028 kg = 515.2 m/s
________________________
For CO2 µ =
√(3 x 8.314 x 298 K)/ 0.044 kg = 411 m/s
_____
d.) r1/r2 = √M2/M1
_______
rx/roxygen = 1.41 =
(1.41)2
√32g/Mx
= 32g = 1.98; Mx = 16g => CH4 gas
Mx
10. If the atmospheric pressure is 0.975 atm, what is the pressure of the
Enclosed gas in each of the cases below?
h
h
h
(I) h = 45 mm
(closed)
(II) h = 78 mm
(open-end)
10. 0.975 atm = 741 mmHg
a.) Pgas = 45mmHg
b.) Pgas = 741 + 78 = 819 mmHg
c.) Pgas = 741 – 67 = 674 mmHg
(III) h = 67 mm
(open-end)
Chapter 5 Some Ideal Gas problems
1. Perform the following conversions:
a) 0.357 atm to torr
b) 6.60x10-2 torr to atm c) 745 torr to
pa d) 802 mmHg to atm
2. A barometer reads the atmospheric pressure 764.7 torr. A
sample of gas is placed in vessel attached to an open-end mercury
monometer. The level of mercury in the open-end arm of the
monometer has a height of 136.4 mm, and that in the arm that is in
contact with the gas has a height 103.8 mm. What is the pressure
of gas in atmosphere?
3. Write an equation or proportionality expression that expresses
each of the following statements:
a) Law of combining volumes explained by Avogadro,s hypothesis.
b) For a given quantity of gas, the product of pressure and
volume is
proportional to absolute temperature.
c) For a given quantity of gas at constant temperature, the
product of pressure and volume is constant.
4. Calculate the following quantities for an ideal gas:
a) the pressure, in atms, if 8.25x10-2 mol occupies 174 ml at -15
0
C;
b) the number of moles in 2.50 L at 37 0C and 725 torr.
c) the volume of a gas occupied by 6.72x10-3 mol at 145 0C and a
pressure of 59.0 torr;
d) the absolute temperature of the gas at which 9.87x10-2 mol
occupies 164 ml at 682 torr;
5. Consider 5.62 L chlorine gas, Cl2 ,at 740 torr and 33 0C.
a) What volume will the gas occupy at 107 0C and 680 torr?
b) What volume will the gas occupy at STP?
c) At what pressure will the volume be 5.00 L if the temperature is 67 0C?
d) At what pressure will the volume be twice as much as original, if
temperature remains constant?
e) How many moles Cl2 gas are in original condition?
6. Perform the following calculations;
a) the density of SO3(g) at 0.96 atm and 35 0C.
b) the density of SO3(g) at STP condition.
c) the molar mass of a gas if its density is 3.67 g/L at 15 0C and 825 torr.
d) the molar mass of gas if 4.40 g occupies 3.50 L at 560 torr and 41 0C.
e) the molar mass of a gas if its density is 2.86 g/L at STP and identify the
molecular formula of the gas.
7. Consider the following reaction of ammonia gas and oxygen gas; at 850 0C
and 5.00 atm.
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
a) How many liters NH3(g) at 850 0C and 5.00 atm are required to react with
1.00 mol of O2(g) in this reaction?
b) How many moles NO(g) can be prepared using 450 L of NH3(g) at a
pressure of 5.00 atm and a temperature of 295 K?
c) How many grams of NH3(g) are needed to generate 10.0 L of NO(g) if
pressure of NO gas is 740 torr at 23 0C?
8. Perform the following calculations;
a) the partial pressure and total pressure of each of the following gases in
the mixture of 4.00 g of each gas CH4(g), C2H4(g) and C4H10(g), in 1.50 L
flask at 0 0C.
b) the partial pressure of each component in a mixture of gases contains
0.55 mol N2, 0.20 mol O2, and 0.10 mol CO2. The total pressure of the
mixture is 1.32 atm.
c) the partial pressure of each component in a mixture of gases contains
3.50 g N2, 2.15 g H2, and 5.27 g of NH3. If the total pressure of the
mixture is 2.50 atm.
9. Consider the following gases at 25 0C; CO2, He, Cl2, CO, CH4, N2
a) place them in order of increasing average molecular speed.
b) calculate and compare the rms speeds of CO(g) and CO2(g).
c) calculate the rate of diffusion of gases CH4(g) and He.
d) which gas effuses 1.41 times faster than O2(g)?
e) which two gases have the same rate of diffusion?
10. If the atmospheric pressure is 0.975 atm, what is the pressure of the
Enclosed gas in each of the cases below?
h
(I) h = 45 mm
(closed)
h
(II) h = 78 mm
(open-end)
h
(III) h = 67 mm
(open-end)