C20 Unit 02 - Gases

Chemistry 20 – Unit 2 – Gases Lesson 1: Introduction By the end of this Unit you will be able to: – explain molecular behaviour, using models of the gaseous state of matter.

By the end of this Lesson you will be able to: – express pressure in a variety of ways, including units of kilopascals, atmospheres and millimetres of mercury. 20–B1.4k – perform calculations, based on the gas laws, under STP, SATP and other defined conditions. 20–B1.4k – explain that the goal of science is knowledge about the natural world 20–B1.2sts

Chemistry 20 – Unit 2 – Gases - The three states of matter include solid, liquid and gas. - The states of matter can be differentiated by the following:

State Solids

Characteristics Definite shape and definite volume

Liquids

Varied shape and definite volume

Gases

Varied shape and varied volume

*

Gases in the Atmosphere - Gases in our atmosphere perform a major function in sustaining life on our planet. The atmosphere is needed for: - breathing and cellular respiration - C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g) - photosynthesis in plants - 6 CO2(g) + 6 H2O(g) → C6H12O6(s) + 6 O2(g) - protection from harmful solar radiation such as ultraviolet light (UV rays) - transmission of heat and water vapour throughout the earth

*

Composition of Gases in the Atmosphere

Gas nitrogen ( N2(g) ) oxygen ( O2(g) ) argon ( Ar(g) ) carbon dioxide ( CO2(g) ) neon ( Ne(g) ) helium ( He(g) ) krypton ( Kr(g) ), xenon ( Xe(g) ), methane ( CH4(g) ), hydrogen ( H2(g) ) nitrogen oxides ( NO2(g), NO3(g) ) sulfur oxides ( SO2(g), SO3(g) ) water vapour

Percentage By Volume 78.1 20.9 0.93 0.03 0.002 0.0005 trace amounts trace amounts trace amounts trace amounts

*

- Although nitrogen and oxygen gases make up about 99 % of our atmosphere, other gases are a source of pollution. For example: - SO2(g) and SO3(g) are called sulfur oxides and contribute to acid rain. - SO2(g) + H2O(g) → H2SO3(aq) sulfurous acid - NO(g) and NO2(g) are called nitrogen oxides and contribute to acid rain. - NO(g) + H2O(g) → HNO2(aq) nitrous acid - CO(g) is given off in incomplete hydrocarbon combustion reactions.

*

Properties of Gases - Under normal conditions, over 99.9 % of the total volume of a gas is empty space, therefore gases can be easily compressed. - Because gas particles or molecules move continuously in a straight line until they hit another object, they spontaneously fill the entire space available. This process can be termed diffusion. - When gas particles collide, no energy is lost so their collisions are said to be elastic. - The higher the temperature of a gas, the more kinetic energy it has and therefore the faster it can move.

*

Gas Pressure - Pressure is defined as a force exerted on an area. - A force is a push or pull and is measured in newtons (N). - Area is measured in square metres (m2). = N2 = 1 Pa ( pascal ) - Pr essure = Force Area m

- When gas particles hit a wall or obstacle, they collectively exert a force (push) on that surface. - If the force is great enough the surface gives way, such as in a balloon, when air forces the material to expand. - If the surface does not give way, say as in a canister, gas pressure builds up. - At sea level, it has been determined that the average pressure of the atmosphere is 101.325 kPa. This value is known as standard pressure or 1.0 atmosphere. *

- Atmospheric pressure is measured by a barometer. - The barometer was invented by Evangelista Torrecelli. He created a barometer utilizing liquid mercury in a long, inverted glass tube (test tube) placed in a pan of mercury. - The height of the mercury would rise or fall depending on the atmospheric pressure changes. If the pressure dropped, the level of mercury would go down. - The average air pressure at sea level is 101.325 kPa, which means that the height of the mercury would be 760 mm, also known as 760 torr. - Therefore: 101.325 kPa = 760 mm Hg = 760 torr = 1 atm

*

Mercury Barometer

760 mm of Hg is 1.0 atmosphere of pressure

*

Conversions 1) 99.0 kPa = _________ mm Hg  760 mm Hg  99.0 kPa ×   = 743 mm Hg  101.325 kPa 

2) 700 mm Hg = ___________ kPa 700 mm Hg ×

101.324 kPa = 93.3 kPa 760 mm Hg

3) 720 mm Hg = ____________ atm 720 mm Hg ×

1.0 atm = 0.947 atm 760 mm Hg

4) 1.43 atm = _____________ kPa 1.43 atm ×

101.325 kPa = 145 kPa 1 atm

*

The Aneroid Barometer - Because the mercury barometer has its disadvantages to its portability an aneroid barometer was developed. - This barometer has a vacuum canister that expands or contracts as the pressure changes. - The change in the shape of the canister causes a needle to move along a scale that reads the pressure. - Very sensitive barometers measure altitude and are called altimeters.

*

The Four Gas Law Variables: Volume, Temperature, Pressure and Moles Volume - All gases must be enclosed in a container. - The three-dimensional space enclosed by the container walls is called volume ( V ). - Volume in chemistry is usually measured in litres ( L ) or millilitres ( mL ). A litre is also called a cubic decimetre ( dm3 ). - A volume of 1 mL = 1 cm3.

*

Temperature - All gases have a temperature, usually measured in degrees Celsius ( °C ), named after Anders Celsius. When the generalized variable of temperature is discussed, the symbol T is used. - There is another temperature scale that is very important in gas behavior. It is called the Kelvin scale ( K ). Note that K does not have a degree sign and Kelvin is capitalized because this was a person's title (Lord Kelvin, his given name was William Thomson). - All gas law problems must be done in Kelvin temperatures. - Standard Temperature is defined as 0 ° C or 273.15 K.

*

Pressure - The molecules of gas hitting the walls of the container create gas pressure ( P ). - There are three different units of pressure used in chemistry: 1. atmospheres ( atm ) 2. millimetres of mercury ( mm Hg ) 3. pascals ( Pa ) or kilopascals ( kPa ) - Standard Pressure is defined as one atm or 760.0 mm Hg or 101.325 kPa. - Standard Temperature and Pressure is abbreviated to STP. Amount of Gas - The amount of gas present is measured in moles ( mol ) or in grams ( g ) . *

Boyle’s Law - Robert Boyle investigated the relationship between pressure and volume in gases, while keeping the temperature constant. - Boyle’s Law states that the volume of a quantity of gas, at a specific temperature, is inversely proportional to the pressure applied to the gas. - This means that if the volume of a gas increases the pressure decreases. - For example, if you had a syringe filled with air, there would be a certain amount of pressure inside the syringe. If you compressed the plunger, the volume of the gas would decrease, but the air pressure inside the cylinder would increase. - Mathematically,

volume is inversely proportional to pressure or 1 Vα P *

- A relationship such as this can be shown to have a constant, k.

PV = k - Inserting values, we can plot the inverse relationship between gas pressure and volume.

Pressure (kPa)

Volume (L)

- As the volume approaches an infinitely small value, gas pressure reaches an infinitely large value. - We can never get to zero volume nor zero pressure. http://www.grc.nasa.gov/WWW/K-12/airplane/aboyle.html

*

Boyle’s Law Problems Using Boyle’s Law, we can calculate the final volume or pressure of a gas, given initial volume and pressure. 1. A container has a volume of 10.0 L and the gas inside exerts a pressure of 75.0 kPa. If the volume of the container was changed to 2.00 L and the temperature was kept the same, calculate the new pressure exerted by the gas in the container. kin the first container = kin the sec ond container k1 = k2 P1V1 = P2V2

( 75.0 kPa ) ( 10.0 L) = ( P2 ) ( 2.00 L) ( 75.0 kPa ) ( 10.0 L) = P 2 ( 2.00 L) 375 kPa = P2 *

2. A syringe’s plunger is withdrawn, changing the volume of the gas from 35.75 cm3 to 83.90 cm3. If the final pressure is calculated to be 57.4 kPa what was the initial pressure of the gas, assuming that the temperature had not changed. kin the first container = kin the sec ond container k1 = k2 P1V1 = P2V2

( P1 ) ( 35.75 cm 3 ) = ( 57.4 kPa ) ( 83.90 cm 3 ) P1

( 57.4 kPa ) ( 83.90 cm 3 ) =

( 35.75 cm ) 3

P1 = 135 kPa

*

Charles’ Law - J.A. Charles formulated a relationship between gas volume and the temperature, while keeping the pressure constant. - He stated that the volume of a gas is directly proportional to the temperature of the gas, assuming the pressure remains constant. - This means that as the volume of a gas decreases or gets smaller, the temperature of the gas also gets smaller. - For example, if you blow up a balloon to a certain volume at room temperature and then expose the balloon to cold winter temperatures, the balloon's volume will decrease, just as the temperature had decreased.

*

volume is directly proportional to temperature or Vα T V - A relationship such as this can be shown to have a constant, k. =k T - Inserting values, we can plot the direct relationship between gas volume and temperature. - As the volume approaches an infinitely small value, temperature, which is measured in Kelvins, reaches an infinitely small value. Temperature - Mathematically,

(K)

- We can never get to zero volume nor zero temperature.

Volume (L)

*

- A Kelvin degree is just another way of measuring temperature and is used in chemistry and physics for very low temperatures. - The lowest possible temperature is at a point where no kinetic energy exists and therefore no molecular movement exists. - This temperature is known to be at zero Kelvin (0 K) and is called absolute zero. ( Note: this is not written 0° K ) - Absolute zero is at –273.15 °C, therefore to measure the temperature in degrees Kelvin we use the formula: K = ° C + 273.15 -To calculate the temperature in °C we use the formula: ° C = K – 273.15

*

100 ° C

0° C

– 273.15 ° C

373.15 K

273.15 K

0K

Boiling Point of Water

Freezing Point of Water

Absolute Zero

*

Temperature Conversions Converting Kelvin to Celsius 25.0 °C = __________ K

25.0 K = ___________ °C

K = °C + 273.15

°C = K − 273.15

K = 25.0 + 273.15

°C = 25.0 − 273.15

K = 298.2 K

°C = −248.2 °C

for significant digits apply adding rule

- Note: For all gas problems that involve temperatures the unit must be in Kelvins.

*

Charles’ Law Problems 1. A certain gas occupies a volume of 35.5 L at a standard temperature of 0.00 °C. What volume will it occupy at 63.78 °C? V1 = 35.5 L Given Information

T1 = 0.00 °C or 273.15 K V2 = ? T2 = 63.78 °C or 336.93 K V1 V2 = T1 T2

35.5 L V2 = 273.15 K 336.93 K

( 35.5 L) ( 336.93 K ) = V 2 ( 273.15 K ) 43.8 L = V2

*

2. A gas has a temperature of 67.4 °C when it is at a volume of 864 L. What will be its temperature if the volume is condensed to 10.0 L?

V1 = 864 L Given Information

T1 = 67.4 °C or 340.6 K V2 = 10.0 L T2 = ? V1 V2 = T1 T2 864 L 10.0 L = 340.6 K T2

( 10.0 L ) ( 340.6 K ) T2 = ( 864 L) T2 = 3.94 K

*

Gay-Lussac’s Law - Gives the relationship between pressure and temperature when volume and amount are held constant. - If the temperature of a container is increased, the pressure increases. - The mathematical form of Gay-Lussac's Law is: P1 P2 = T1 T2

*

Problems 1. 10.0 L of a gas is found to exert 97.0 kPa at 25.0 °C. What would be the required temperature (in Celsius) to change the pressure to standard pressure? V = 10.0 L V = 10.0 L 1

P1 P2 = T1 T2

P1 = 97.0 kPa T1 = 25.0 °C

2

PP22 == 101.325 kPa T2 = ? °C

( 97.0 kPa ) = ( 101.325 kPa ) ( 298.2 K ) T2 ( 101.325 kPa )( 298.2 K ) T2 = ( 97.0 kPa ) T2 = 311 K T2 = 311 K − 273.15 °C T2 = 38 °C *

2. 5.00 L of a gas is collected at 22.0 °C and 745.0 mm Hg. When the temperature is changed to standard, what is the new pressure? V V11 == 5.00 L

V2 = 5.00 L

TT11 == 22.0 °C

TT22 == 273.15 K

PP11 == 745.0 mm Hg

P P22 == ?

P1 P2 = T1 T2

( 745.0 mm Hg ) = ( P2 ) ( 295.2 K ) ( 273.15 K ) ( 745.0 mm Hg )( 273.15 K ) = P 2 ( 295.2 K ) 689 mm Hg = P2

*

Avogadro's Law

- Gives the relationship between volume and amount when pressure and temperature are held constant. Amount is measured in moles. - If the amount of gas in a container is increased, the volume increases. - Suppose the amount is increased. This means there are more gas molecules and this will increase the number of impacts on the container walls. This means the gas pressure inside the container will increase, becoming greater than the pressure on the outside of the walls. This causes the walls to move outward. Since there is more wall space the impacts will lessen and the pressure will return to its original value. V1 V2 - The mathematical form of Avogadro's Law is: = n1 n2 - This means that the volume-amount fraction will always be the same value if the pressure and temperature remain constant. *

Problem: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)? V1 V2 = n1 n2

( 5.00 L)

V2 = ( 0.965 mol ) ( 1.80 mol )

( 5.00 L)( 1.80 mol ) = V 2 ( 0.965 mol ) 9.33 L = V2

*

Combined Gas Law To derive the Combined Gas Law, do the following: Step 1: Write Boyle's Law: Step 2: Multiply by Charles Law:

Step 3: Multiply by Gay-Lussac's Law:

P1V1 = P2V2 V1 V2 P1V1 × = P2V2 × T1 T2 P1V12 P2V22 = T1 T2 P1 P1V12 P2 P2V22 × = × T1 T1 T2 T2 P12V12 P22V22 = 2 T1 T22

Step 4: Take the square root to get the combined gas law:

P1V1 T1

=

P2V2 T2

*

Problems 1. 2.00 L of a gas is collected at 25.0°C and 745.0 mm Hg. What is the volume at STP? PP11 == 745.0 mm Hg P2 = 760 mm Hg V V11 == 2.00 L

V2 = ?

TT11 == 298.2 K

TT22 == 273.15 K P1V1 P2V2 = T1 T2

( 745.0 mm Hg )( 2.00 L) = ( 760 mm Hg )( V2 ) ( 298.2 K ) ( 273.15 K ) ( 745.0 mm Hg )( 2.00 L)( 273.15 K ) = V 2 ( 298.2 K )( 760 mm Hg ) 1.80 L = V2 *

2. A container is reduced to a volume of 125 cm3 when a temperature of –75.00 °C is reached. If the pressure changes from 1.20 atm to 852 torr and the initial temperature was 20.00 °C, calculate the initial volume. P11 = 1.20 atm

PP22 == 852 torr

V1 = ?

V22 = 125 cm 3

T11 = 20.00 °C

TT22 == −75.00 °C

P1V1 P2V2 = T1 T2

( 912 torr )(V1 ) ( 293.15 K )

( 852 torr ) ( 125 cm 3 ) = ( 198.15 K )

( 852 torr ) ( 125 cm 3 ) ( 293.15 K ) V1 = ( 198.15 K )( 912 torr ) V1 = 173 cm 3

*

Molar Volume - The molar volume is the volume occupied by one mole of ideal gas at STP or SATP (RTP). - It is the volume of ANY ideal gas at Standard Temperature and Pressure (STP) or Standard Ambient Temperature and Pressure (SATP or RTP). - Its value is: STP : 22.4

L mol

: T = 0 ° C and P = 1 atm

L SATP ( RTP ) : 24.8 mol

: T = 25 °C and P = 1 atm

*

Example #1: You have 2.00 L of dry H2 at STP. a) How many moles is this? b) What is the mass of the hydrogen gas? Solution: a) If 22.4 L of gas equals 1 mol of gas, we can set up a simple ratio: 22.4 L 2.00 L = 1 mol x mol 22.4 x = 2.00 x = 0.0893 mol b)

m = nM  m = ( 0.0893 mol )  2.02 m = 0.180 g



g   mol 

*

Example #2: 0.250 mol of HCl(g) will occupy how many litres at STP? Solution:

22.4 L xL = 1 mol 0.250 mol

x = 5.60 L Example #3: What is the molar volume at 576 K? Solution: Since 576 K is not a standard temperature, we must create a ratio to find the new volume. L   L    22.4  x  mol   mol   = 273.15 K 576 K 12902.4 = 273.15 x L 47.2 =x mol *

Ideal Gas Law - We can combine all gas laws into one convenient expression: where

PV = nRT P = pressure in kPa

V = volume in L n = moles of gas T = temperature in K R = Universal Gas Constant = 8.314 kPa⋅ L⋅ mol –1⋅ K –1 or kPa ⋅ L    8.314  mol ⋅ K  

*

Example #1: What mass of hydrogen gas can be produced if the gas occupies a volume of 50.0 mL at 20.00 °C and 94.5 kPa? Given Information: m= ? V = 50.0 mL or 0.0500 L T=

20.00 ° C or 293.15 K

P=

94.5 kPa

To solve for mass we need the formula: m = nM

*

Since number of moles is unknown, we need to use the given information to solve for moles. PV = nRT PV n= RT

n=

( 94.5 kPa ) ( 0.0500 L) kPa ⋅ L   8 . 314   ( 293.15 K ) 

mol ⋅ K 

n = 1.94 x 10 − 3 mol Now, calculate for mass: m = nM g   m = ( 1.94 x 10 − 3 mol )  2.02  mol   m = 0.00392 g *

Example #2: A balloon has a volume of 10 000 L at 100 kPa and 20.00 °C. If one was to fill this balloon with helium, how many moles of helium would be required? Given:

V = 10 000 L P=

100 kPa

T=

20.00 ° C or 293.15 K

n=

?

PV = nRT PV n= RT n=

( 100 kPa ) ( 10000 L) kPa ⋅ L   8 . 314   ( 293.15 K ) 

mol ⋅ K 

n = 410 mol *

Example #3: A storage tank contains 2.65 kg of methane gas at 735.2 kPa and 25.00 °C. What is the tank’s volume? Given:

m = 2.65 kg or 2.65 x 10 3 g P = 735.2 kPa T = 25.00 ° C or 298.15 K V= ?

m M ( 2.65 × 10 3 g ) n= g   16 . 05   mol   n = 165 mol n=

PV = nRT nRT V= P kPa ⋅ L   ( 165 mol )  8.314  ( 298.15 K ) mol ⋅ K   V= ( 735.2 kPa ) V = 557 L *

Example #4: If 38.5 g of nitrogen gas is found in a container with a volume of 15.27 L and 1.19 atm, what is the temperature in degrees Celsius? Given:

n= n=

m= V= P= T=

38.5 g 15.27 L 1.19 atm or 121 kPa ?

m M

( 38.5 g ) g   28 . 02   mol  

n = 1.37 mol

PV = nRT PV T= nR

( 121 kPa ) ( 15.27 L ) T= ( 1.37 mol )  8.314 kPa ⋅ L  T = 161 K



mol ⋅ K 

T = −112 °C *

Dalton’s Law of Partial Pressure - John Dalton’s law of partial pressure states that the total pressure exerted by a mixture of gases is the sum of the partial pressures of each gas in the container. Ptotal = Pgas 1 + Pgas 2 + Pgas 3 + etc Example: A container has Gas A at a pressure of 1.25 atm, Gas B at a pressure of 102 kPa and Gas C at a pressure of 720 mm Hg. What is the total pressure, in kPa, inside the container?

Ptotal = PGas A + PGas B + PGas C

Ptotal = 127 kPa + 102 kPa + 96.0 kPa Ptotal = 325 kPa

*

Example: The total pressure inside a container is 140 kPa. If 7 molecules of Gas A, 2 molecules of Gas B and 1 molecule of Gas C exist, calculate the pressure exerted by each gas. Total molecules = 10 molecules Gas A: 7 molecules or 70 % Gas B: 2 molecules or 20 % Gas C: 1 molecule or 10 % Gas A: 70 % of 140 kPa = 98.0 kPa Gas B: 20 % of 140 kPa = 28.0 kPa Gas C: 10 % of 140 kPa = 14.0 kPa

*

Ideal Gases vs Real Gases - We have assumed that all gases obey the Ideal Gas Law ( PV = nRT ). - In an “Ideal” gas, molecules have no volume and no intermolecular forces. This assumption must be made in order to make the Ideal Gas Law formula work in all cases. - While the Ideal Gas Law does a very good job of predicting gas behavior, it's not perfect. - Real gases differ in two major ways from an ideal gas 1) Particles in a real gas have volume. 2) Particles in a real gas attract each other, especially when they are condensed. *

Summary

STP

SATP

Molar Volume

Temperature

Pressure

273.15 K

101.325 kPa 1.0 atm 760 mm Hg 760 torr

22.4

100 kPa

24.8

298.15 K

L mol

L mol

*

- Boyle's Law:

P1V1 = P2V2 - Charles’ Law:

V1 V2 = T1 T2 - Gay-Lussac’s Law:

P1 P2 = T1 T2 - Avogadro's Law:

V1 V2 = n1 n2

- Combined Gas Law:

P1V1 P2V2 = T1 T2 - Ideal Gas Law:

PV = nRT

L ⋅ kPa - R = 8.314 mol ⋅ K - Dalton’s Law of Partial Pressure:

Ptotal = Pgas 1 + Pgas 2 + Pgas 3 + etc

*

Gas Law Stoichiometry - In our stoichiometry unit, we looked at four different types of questions: Mole to Mole

Mole to Mass

Mass to Mole

Mass to Mass

Given – moles

Given – moles

Given – Mass

Given – mass

Required moles

Required mass

Required moles

Required mass

*

- To solve these questions, we look at our given information, looked at our required information and then decided on the steps to take: MASS of GIVEN

MASS of REQUIRED

?

mG nG = MG

m R = nR M R

Moles of given nG

Moles of required nR

*

- Now, we are going to include volume to stoichiometry questions. We use the Ideal Gas Law to do this: MASS of GIVEN

MASS of REQUIRED

?

mG nG = MG

m R = nR M R Moles of given nG

VR P nR = RT VOLUME of GIVEN

Moles of required nR

nR RT VR = P VOLUME of REQUIRED

*

Problems 1. What volume of CO2(g) is produced at 94.5 kPa and 115.5 °C in the exhaust of a car if 100 g of gasoline (C8H18(g)) are burned?

V= ? P = 94.5 kPA T = 115.5º C or 388.7 K mC8 H18 ( l ) = 100 g

*

2 C8 H 18 ( l ) + 25 O2 ( g ) → 16 CO2 ( g ) + 18 H 2O( g )

m = 100 g m n= M n=

nCO2

( 100 g )  114.26 g    mol  

n = 0.875 mol

nCO2

16 = 0.875 mol × 2 = 7.00 mol

V =? nRT V= P kPa⋅ L ( 7.00 mol ) ( 8.314 mol ⋅ K ) ( 388.7 K ) V= ( 94.5 kPa ) V = 239 L *

2. How many moles of sodium must react with H2O(l) to produce 250 mL of H2(g) at 23.6 °C and 102.6 kPa? 2 Na(s)

+

2 HOH(l)

n=?

→

H2(g)

+

2 NaOH(aq)

V = 0.250 L T = 23.6° C or 296.8 K P = 102.6 kPa

nNa( s )

2 = 0.0104 mol × 1

nNa( s ) = 0.0208 mol

PV RT ( 102.6 kPa ) ( 0.250 L) = kPa⋅ L ( 8.314 mol ⋅ K ) ( 296.8 K )

nH 2 = nH 2

nH 2 = 0.0104 mol

*

3. If 17.8 L of hydrogen, at 22.5 °C and 21.4 kPa, react with nitrogen to produce 22.7 L of ammonia at 33.9 kPa, what’s the temperature of the ammonia (in °C)? 3 H2(g)

+

→

N2(g)

2 NH3(g)

V = 17.8 L

V = 22.7 L

T = 22.5 ° C or 295.7 K

P = 33.9 kPa

P = 21.4 kPa

T=?° C

nH 2

PV = RT

nH 2

( 21.4 kPa ) ( 17.8 L ) = ( 8.314 ) ( 295.7 K )

nNH 3 = 0.155 mol ×

kPa⋅ L mol ⋅ K

nH 2 = 0.155 mol

nNH 3 = 0.103 mol TNH 3

2 3

( PV 33.9 kPa ) ( 22.7 L ) = = kPa⋅ L nR ( 0.103 mol ) ( 8.314 mol ⋅K )

TNH 3 = 896 K or 623 °C

*

4. What volume of carbon monoxide gas could be produced at STP by the incomplete combustion of 50.0 kmol of coal (assume pure carbon)? 2 C(s) + O2(g) → 2 CO(g) nC = 50.0 kmol

V=? T = 273.15 K P = 101.325 kPa

VCO( g ) VCO( g )

nRT = P kPa⋅ L ( 50.0 kmol ) ( 8.314 mol ⋅ K ) ( 273.15 K ) = ( 101.325 kPa )

nCO( g ) nCO( g )

2 = 50.0 kmol × 2 = 50.0 kmol

VCO( g ) = 1120636 L VCO( g ) = 1120.636 kL or 1.12 × 10 6 kL

VCO( g ) = 1.12 ML *