Level L Chemistry BQ Chapter 5

8/19/2019 Level L Chemistry BQ Chapter 5

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Chapter 5 - The Gas Phase

List ALL

Basic Questions

Basic Question 1

5.2.2 Measure the pressure of a gas using a closed end manometer -

Find the pressure of the gas in the closed end manometer in each of the following cases.

a)

If the difference between the heights of A andB is 120 mmHg, and the atmospheric pressureis 750mmHg. What is the pressure of gas X in

the manometer?

In a closed end manometer, the pressure of the

gas is equal to the difference in height between

the two sides of the manometer

Pgas = 120 mmHg

b)

If there is no difference between the heights ofA and B, and the atmospheric pressure is 660mmHg. What is the pressure of gas X in the

manometer?

In a closed end manometer, when there is no

difference in height between the two sides of themanometer, this means there is no gas in the

gas bulb

Pgas = zero


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Basic Question 2

5.2.2 Measure the pressure of a gas using an open end manometer -

A diagram for an open-end manometer is shown below.

If the flask is open to the atmosphere, the mercury levels are equal. For each of the following situationswhere a gas is contained in the flask, calculate the pressure in the flask

a) If the difference between theheights of A and B is 150

mmHg, and the atmospheric

pressure is 800 mmHg. What

is the pressure of gas X in themanometer?

P gas > P atm

P gas = P atm + h

= 800 + 150

= 950 mmHg

b) If the difference between theheights of A and B is 120

mmHg, and the atmospheric



P gas < P atm

P gas = P atm - h

= 700 - 120

= 580 mmHg

c) If there is no difference between the heights of A and

B, and the atmospheric



P gas = P atm

= 740 mmHg


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Basic Question 3

0.10 mole of oxygen gas and 0.40 mole of nitrogen gas are placed in an empty container of volume 20

liters. The total pressure in the container is 0.50 atm. Find the:

5.2.3 Find total # of moles given the individual # of moles in a gaseous mixture G

a)

Total number of moles of gas in the container.

Given

n O2 = 0.10 mol

n N2 = 0.40 mol

V = 20 L

Ptotal = 0.50 atm

R.T.F

n total = ????

The total number of moles= n O2 + n N2 = 0.10 + 0.40 = 0.50 moles.

5.2.3 Find mole fractions given the individual # of moles in a gaseous mixture -

b) Mole fraction of nitrogen gas and oxygen gas in the container.

Given

n O2 = 0.10 mol

n N2 = 0.40 mol

V = 20 L

Ptotal = 0.50 atm

R.T.F

= ???= ???

Mole fraction of N2 = 0.40

0.50 = 0.80

Mole fraction of O2 =0.10

0.50 = 0.20


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5.2.3 Find partial pressure given the individual # of moles in a gaseous mixture and Ptotal -

c) Partial pressure of oxygen and nitrogen gases

Given

n O2 = 0.10 moln N2 = 0.40 mol

V = 20 L

Ptotal = 0.50 atm

R.T.F

= ????= ????

2OP = mole fraction of O2 × total pressure = 0.20 × 0.50 = 0.10 atm

2NP = Total pressure -

2

PO

= 0.50 – 0.10 = 0.40 atm

OR

2NP

2N total

= X ×P = 0.80×0.50 = 0.40 atm

5.2.3 Know that every gas in a container occupies the entire container G

d) Volume of the oxygen gas and nitrogen gas in the container

Given

n O2 = 0.10 mol

n N2 = 0.40 molV = 20 L

Ptotal = 0.50 atm

R.T.F

= ????= ????

Volume of O2 = volume of the container = 20 litres

Volume of N2 = volume of the container = 20 litres

5.2.3 Know that fraction of volume occupied by any gas in a container is 100% G

e) Fraction of the volume of the container occupied by oxygen. 100%


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Basic Question 4

A sample of air was found to contain 0.32 g oxygen and 1.12 g nitrogen. Find the:

5.2.3 Find total # of moles given the masses of individual gases in a mixture G

a) Total number of moles of gas in the sample.

Given

m O2 = 0.32 g

m N2 = 1.12 g

R.T.F

= ? ? ? ?

Total number of moles = number of moles of O2 + number of moles of N2

Total number of moles =0.32 1.12

+ = 0.010 + 0.040 = 0.050 moles32 28

5.2.3 Find mole fractions given the masses of individual gases in a mixture -

b) Mole fraction of oxygen in the sample.

Given

m O2 = 0.32 g

m N2 = 1.12 g

R.T.F

= ????

Mole fraction of oxygen = number of moles of O2/total number of moles =

0.3232

= 0.200.050

5.2.3 Find partial pressures given the masses of individual gases in a mixture and Ptotal -

c) Partial pressure of each of oxygen and nitrogen in the sample; knowing that the sample of air was

exerting a pressure of 755 mmHg.

Given

m O2 = 0.32 g

m N2 = 1.12 g

Patm = 755 mmHg

R.T.F

= ????= ????


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2OP = mole fraction of O2× total pressure = 0.20×755 = 151 mmHg

2NP = 755 -151 = 604 mmHg or

2 2N N Total

1.1228P = X × P = ×755 = 604 mmHg

0.050

5.2.3 Find % composition of air given the masses of O2 and N2 gases in a mixture -

d) The percentage composition of air.

% by mass O2 =0.32

×100 =0.32 + 1.12

22.2%

% by mass N2 = 77.8%

All Applications on the gas laws

Basic Question 5

5.3.4 Application on P1V1 = P2V2 TG

A certain mass of helium gas was placed in a cylinder fitted with a piston at 25C. The volume and pressure were measured to be 3.0dm

3 and 120 mmHg respectively.

a) If the piston was pushed till the volume became 2.0dm3 while maintaining a constant temperature,

how much will the pressure be?

Given

P1 = 120 mmHg

V1 = 3.0dm3

V2 = 2.0dm3

n and T are constant

R.T.F

= ????

Since n and T are constant then PV = ct

P1V1 = P2V2.

(120)(3.0) = (P2)(2.0)P2 = 180 mmHg


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5.3.4 Application on =

TG

b) If the gas is heated to 125C, while maintaining the volume at 3.0dm3, how much will the pressure

be?

Given

P1 = 120 mmHg

t1 = 25 C

t2 = 125 C

n and V are constant

R.T.F

= ????

Since n and V are constant then= ct

=

+ = + P2 = 160 mmHg


TG

c) If the gas is heated to 125C, the pressure was found to become 90 mmHg, how much will the

volume be?

Given

P1 = 120 mmHg

t1 = 25 C

t2 = 125 C

V1 = 3.0 dm3

P2 = 90 mmHg

R.T.F

= ????

Since n is constant then = ct =

(.)(+) = ()(+) V2 = 5.3 dm3


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TG

d) If the gas is heated to 125C while keeping the pressure constant, how much will the volume be?

Given

V1 = 3.0 dm3

t1 = 25 C

t2 = 125 C

n and P are constant

R.T.F

= ????

Since n and P are constant then= constant =

.(+) = (+) V2 = 4.0 dm3 Basic Question 6


TG

2.0g of helium gas were placed in a glass container at 25C. The pressure was measured to be 120mmHg. If 6.0g of He gas were added to the container while maintaining the temperature constant, how

much would the pressure be?

Given

m1 = 2.0 g

P1 = 120 mmHg

m2 = 2.0 + 6.0 = 8.0g

T and V are constant

R.T.F

= ????

Since V and T are constant then = constant = n1 =

.. = 0.50 mole n2 = .. = 2.0 moles. = . P2 = 480 atm


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Basic Question 7


TG

2.0g of helium gas were placed in a balloon at 25C. The volume of the balloon was measured to be5.0L. If 6.0g of He gas were added to the balloon while maintaining the temperature and pressureconstant, how much would the volume be?

Given

m1 = 2.0 g

V1 = 5.0L

m2 = 2.0 + 6.0 = 8.0g

T and P are constant

R.T.F

= ????

Since P and T are constant then= constant

= n1 =

.. = 0.50 mole, n2 =.. = 2.0moles

.. = . V2 = 20 L

Basic Question 8

5.3.4 Given V, P (mmHg) and T( C), find # of moles of an ideal gas TG

How many moles of an ideal gas occupy 2.24 dm3 at a pressure of 380 mm Hg and a temperature of

273ºC? (No calculator allowed) Take R =

22.4

273 dm

3.atm.K

-1mol

-1

Given

V = 2.24 L

P = 380 mmHgt = 273 C

R.T.F

n = ????

= = .. () = .


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Basic Question 9

5.3.4 Given m(kg), T( C) and P (mmHg), find volume of an ideal gas TG

When 1.0 kg of He gas is filled into an empty balloon at 27ºC the pressure in the balloon becomes 760

mm Hg. Find the volume of the inflated balloon.

Given

m = 1.0 kg = 1000 g

P = 760 mmHg= 1atm

t = 27 C

R.T.F

V = ????

V = = = 6150 dm3 = 6.150 x 10

3 dm

3

Basic Question 10

A cylinder fitted with a piston has 10 dm3 of a gas 27

ºC. How can you double the pressure by varying

5.3.4 Application on Boyle’s law: 1 1 2 2P V P V G

a) only the volume?

Given

V1 = 10dm

3

P1P2 = 2P1

R.T.F

V2 = ????

P α, for a fixed amount of a gas at constant temperature.

So halving the volume doubles the pressure.

Make the volume 5 dm3.


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5.3.4 Application on 1 12 2

P T

P T

G

b) only the temperature? (No calculator allowed)

Given

t1 = 27 C T1 = 27 + 273 = 300K

P1

P2 = 2P1

R.T.F

T2 = ????

P α T , for a fixed amount of gas at constant volume, doubling the Kelvin temperature will double

the pressure,

Since 1 1

2 2

P T=

P T

Since P 2 =2 P1 Then1 1

1 2

P T=

2 P T and T2 = 2T1 = 2×(27+273) = 600K = 327°C

This means to double the pressure, you need to do one of the following:

1. Double the temperature in Kelvin

2. Increase the temperature by 300C

Basic Question 11

A 6.0 dm3 flask is filled with dry air to 1,000 mmHg pressure. Air is 20 percent O2 and 80 percent N2 by

volume (this means by mole ratio).

5.3.4 Know that a gas occupies the entire volume of the container it is in TG

a) What is the volume of the oxygen in the flask and that of nitrogen?

Volume of oxygen gas = volume of nitrogen gas = volume of the container = 6.0 dm3

5.3.4 Find partial pressure of each gas in a mixture given the % composition and Ptotal -

b) What is the partial pressure of each gas? (No calculator allowed)

Given

V = 6.0 dm3

Pt = 1000 mmHg

Air 20 % O2 and 80 % N2

R.T.F

= ???? = ????

2OP = mole fraction of O2 × total pressure = 0.20 × 1000 = 200mmHg

2NP = mole fraction of N2 × total pressure = 0.80 × 1000 = 800mmHg


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Basic Question 12

5.3.4 Given t( C), P(atm) calculate the molar volume TG

Calculate the molar volume of an ideal gas at 27°C temperature and 1.0 atm.

Given

t = 27°C T = 27 + 273 = 300K

P = 1.0 atm

n = 1 mole

R.T.F

V = ?

V = = = 24.6 dm3 = 25 dm

3

Basic Question 13

5.3.4 Find the average molar mass of air _

Consider air to be 80% N2 and 20 % O2 (No calculator allowed). Calculate:

a) The average molar mass of air

Assume 100 moles of air

There will be 20 moles of O2 and 80 moles of N2.

Total mass of 100 moles of air = mass of 20 moles of O 2 + mass of 80 moles of N2

= (20×32) + (80×28) = 2880g

Average molar mass of air = 2880/100 = 28.8 g/mol

5.3.4 Find the density of air at STP given molar mass of air and its % composition -

b)

The density of air at STP.

The STP conditions are 0°C and 1.0 atm

PM = DRT

PM (1.0)(28.8)D = = = 1.3 g/L

22.4RT(273)

273


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Basic Question 14

5.3.4 Application on PM = dRT _

Calculate density of hydrogen gas if 20g were found to exert a pressure of 760mmHg in a 5.0 dm3

container at a temperature of 127C.

Given

m = 20 g

P = 760 mmHg = 1.0 atm

V = 5.0 dm3

R.T.F

density = ?

PM = DRT

-3PM (1.0)(2.0)D = = = 0.061 g.dm

RT (0.082)(400)

Basic Question 15

5.3.4 Find P of an ideal gas given mass, volume and t( C) TG

Calculate pressure of 16.0 g of oxygen gas in a103 liter container at a temperature of 91°C.

[Hint: 273 = 3 × 91] (No calculator allowed)

Given

m = 16.0 g

V = Lt = 91

°C T = (91 + 273) K

R.T.F

P = ?

16.0 22.4(91+ 273)

nRT 32 273P = = = 4.48 atm

10V

3


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Basic Question 16

5.4 Know that gases mix by diffusion TG

The diagram shows some people sitting round a dinner table.

When the lid of the dish was removed, all the people could smell the food. How did the smell reachthem?

Diffusion

Basic Question 17

A vessel contains equal moles of SO2 and He gases. The pressure is 760 mm Hg.

5.4 Compare the velocities of two given gases at the same temperature -

a) On average, which travels faster, the He atoms or SO2 molecules? How many times as fast?

Given

A vessel containing He and SO2

R.T.F

Compare the speeds of the two gases

Method 1

At same temperature, average KE is the same

2

2

2

2

2 2

He SO

2 2

He SO

2 2

He SO

2 2

He SO

He SO SO

KE = KE

1 1Mv = Mv

2 2

1 1× 4× v = × 64× v

2 2

v = 16v

v = 16 v v

Helium travels faster, 4 times faster than SO2 .


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Method 2

At same temperature, average KE is the same

=

= = √ = Helium travels faster, 4 times faster than SO2.

5.4 Given the total pressure in a vessel containing equal moles of two

gases, predict the pressure if one of the gases is removed

-

b) If the SO2 was removed from the system, what will the pressure become?

Given

n of SO2 = n of He

same V and T

R.T.F

P = ?????

In the same vessel under the same temperature, Pα n.

Since equal number of moles are present, if we remove SO2 from the vessel, we are halving the

number of moles in the vessel so pressure halves as well. Pressure becomes 380mmHg.

5.4 Know the effect on KE of increasing/decreasing temperature -

c) What happens to the kinetic energy of the molecules if the temperature is raised from 0°C to 273

°C?

Given

T1 = 0 + 273 = 273K

T2 = 273 + 273 = 546K

R.T.F

Compare KE1 and KE2

Since K.E.α T

2

1

2

1

KE T (273+ 273)= = = 2

KE T (0 +273)

So the K.E. doubles.


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Sample Questions

I n all the mul tiple choice questions, more than one answer could be corr ect.

5.1 Molar volumes of gases

5.1.1 The volumes occupied by a mole of nitrogen

Sample Question 1

Know that the molar volume in the gaseous state is much larger than the liquid state

The volume of one mole of nitrogen in the liquid state is about 27 cm3. About how many times is the

volume in the gaseous state (at STP) larger?

a) 1 time (same size) b) 10 times

c) 100 times

d) 1,000 times

e)

1,000,000 times

Sample Question 2

Know the meaning of the volume of a gas

When we talk of the volume of a certain sample of gas we mean

a) the volume of one mole. b) the volume of one molecule.

c) the sum of volumes of all the molecules in the sample.

d) the volume of the empty container

e)

the volume of the gas independent of the container.

5.1.2 Molar volumes of other gases at STP

Sample Question 3

Know how the molar volume of gases changes with molar mass

Measurement of molar volumes of various gases show that the molar volume

a) tends to decrease with increasing molar mass.

b) tends to increase with increasing molar mass.

c)

is totally independent of molar mass.d) is proportional to the molar mass.


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Sample Question 4

Know how the molar volume of gases changes with atomicity

Measurement of molar volumes of various gases show that the molar volume

a) tends to decrease with increasing atomicity. b) tends to increase with increasing atomicity.

c) is totally independent of atomicity.

d)

is proportional to the atomicity.

Sample Question 5

Given mass and volume of gas at STP, find mass of 1 mole

5.6 dm3 of a gas at STP have a mass of 8.0 g. The mass of one mole of this gas is (no

calculator allowed)a) 44.8 L

b) 44.8 g

c) 8 L

d)

8 ge) 32 g

Steps:

1 mole of a gas at STP occupies 22.4 dm3

5.6 dm3 8.0 g

22.4 dm3 ??? 32 g

5.1.3 PV behavior of a real gas

5.1.4 The meaning of “volume of a gas”

5.2 The Kinetic Theory of Gases

5.2.1 Temperature and the kinetic theory

Effect of increasing the temperature on the product P ×V

Sample Question 6

Kinetic theory of gases

Describe, in one sentence, the motion of a gas, as pictured by the postulates of the kinetic theory.

The tiny particles of a gas are moving at random, colliding with each other and with the walls of

the container.


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Sample Question 7

Know why at a higher temperature a gas exerts a higher pressure

According to the kinetic theory, at a higher temperature a gas exerts a higher pressure because

a) its particles collide with the walls of the container more frequently.

b) its particles collide with the walls of the container more strongly.c) it will have more molecules.

Sample Question 8

Know that at a higher temperature gas molecules move more rapidly

According to the kinetic theory, at a higher temperature the molecules of a gas

a) move more rapidly.

b) move more slowly.c) become lighter.

Kinetic energy and temperature

Sample Question 9

Know that at same temperature different gases have the same molecular KE

According to the kinetic theory, the molecules of the oxygen gas have, at the same temperature,

a) the same average speed as hydrogen molecules.

b) the same average kinetic energy as hydrogen molecules.c) the same mass as hydrogen molecules.

Effect of temperature on the volume of a gas

Sample Question 10

Know effect of temperature on volume of gas at constant P

The temperature of a fixed amount of gas is changed at a constant pressure, every time taking the

readings. When the volume (on the ordinate) is plotted against the temperature (abscissa) the graph isa) a curve that passes through the point t = 0°C

b) a straight line which when extrapolated backward passes through the point t = 0°C

c) a straight line which when extrapolated backward passes through the point t = −273°C


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Sample Question 11

Know what is meant by an ideal gas

What is an ideal gas?

An ideal gas is a gas that does not liquefy

The molecules of an ideal gas have zero volume

The molecules of an ideal gas have no forces of attraction between them.

The absolute temperature scale

Sample Question 12

Know that the absolute zero is 0K which is −273°C

What is the temperature that is called the absolute zero?a) It is the temperature at which liquid water becomes solid ice.

b) It is the temperature of – 273.16 °C, or 0 K.

c) It is the temperature at which air becomes a liquid.

Sample Question 13

Know the magnitude of the Kelvin

What is the magnitude (size) of the Kelvin as compared to a °C?

a) The size of the Kelvin is the same size as the Celsius degree: 1K = 1°C.

b) The size of the Kelvin is the same – 273.16 °C.c) The size of the Kelvin is the same size as Fahrenheit degree.

Sample Question 14

Changing Celsius to Kelvin and vice versa

a) Change 373 K to °C. T = t℃ + 273 , 373K – 273 = 100°C b) Change 27°C to Kelvin. 27°C + 273 = 300K

Sample Question 15

Volume of a gas is directly proportional to the absolute temperature

Which relation is true between the volume of a fixed amount of gas and its temperature, at

constant pressure?a) As the temperature increases the pressure decreases. b) The volume is directly proportional to the Celsius temperature.

c) The volume is directly proportional to the Kelvin temperature.


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Sample Question 16

Relation between FP and BP of a gaseous substance and its molar mass

How do the boiling points and freezing points in degrees Celsius of certain substances that are gaseous

at room temperature change with increasing molar mass?

a) In general, the higher the molar mass the higher is the FP and BP. b) In general, the higher the molar mass the lower is the FP and BP.

c)

In general, the freezing points and boiling points are directly proportional to the molar mass.

5.2.2 Measuring pressure

The barometer

Sample Question 17

The barometer is used to measure atmospheric pressure

The barometer is used to measure

a) absolute temperature.

b)

Distance.c) the pressure inside a tyre.

d) the pressure of a gas in a bottle.

e) the atmospheric pressure.

Sample Question 18

Know what the unit ‘atmosphere’ means

The atmosphere is the pressure that

a) is equal to 1 newton/m2.

b)

can support a column of mercury 1.00m high at 0°C.c) can support a column of mercury 760 mm high at 0°C.

d) can support a column of mercury of volume 22.4 dm3 at 0°C.

e) can support a column of mercury of 50 km high at 0°C.

f) can support a column of mercury 76 cm high at 0°C.


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The closed-end manometer

Sample Question 19

Know when and how to use the closed-end manometer

I. A closed-end manometer is used to

a) measure directly the pressure of a gas in a container. The pressure of the gas in the

container is measured by the difference in height of mercury between the two arms of themanometer.

b) measure the atmospheric pressure. The pressure reading is taken as the level of the mercury.

c) measure the difference between the pressure of a gas in a container and the atmosphere. The pressure of the gas in the container is measured by the difference in height of mercury between

the two arms of the manometer.

II. Which of the following is true regarding the closed-end manometer?a) The closed-end manometer should have air in the sealed end.

b) The level of the mercury in the closed-end side can never be lower than the level of the

mercury in the other side.

c)

If the gas container is empty (i.e. it has vacuum), the level of mercury on both sides of thetube is the same.

d) The difference in height of mercury between the two arms of the manometer is independent

of the atmospheric pressure.

Sample Question 20

Determine the pressure in a flask using a closed-end manometer.

A flask containing some gas is connected to a closed-end manometer. The atmospheric pressure is 758

mm Hg. Which of the following is correct, and which is wrong?

gas

arm A arm B

a) If the mercury level in A is 302 mm lower than in B, the pressure of the gas is 302 mm Hg.

Correct b) If the mercury level in A is 302 mm lower than in B, the pressure of the gas is 1,060 mm Hg.

Wrong c) If the mercury level in A is 302 mm higher than in B, then some gas must have leaked into the

closed end, and the manometer does not function properly. Correct


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The open-end manometer

Sample Question 21

Know when and how to use the open-end manometer

An open-end manometer is used to

a) measure directly the pressure of a gas in a container. The pressure of the gas in the container is

measured by the difference in height of mercury between the two arms of the manometer. b) measure the atmospheric pressure. The pressure reading is taken as the level of the mercury.

c) measure the difference between the pressure of a gas in a container and the atmosphere.

The pressure of the gas in the container is measured by the difference in height of mercury

between the two arms of the manometer.

Sample Question 22

Determine the pressure in a flask using an open-end manometerA flask containing some gas is connected to an open-end manometer.

The atmospheric pressure is 758 mm Hg. Which of the following is correct?

gas

Arm A arm B

a) If the mercury level in A is 302 mm lower than in B, the pressure of the gas is 302 mm Hg.

b) If the mercury level in A is 302 mm lower than in B, the pressure of the gas is 1,060 mm

Hg.

c) If the mercury level in A is 302 mm higher than in B, the pressure of the gas is 456 mm Hg.

5.2.3 Partial pressure

Sample Question 23

Know the meaning of partial pressure of a gas in a mixture of gases

The partial pressure of a gas in a mixture of gases placed in a vessel is:

a) The ratio of the pressure exerted by this gas to the total pressure exerted by the mixture.

b) The total pressure diminished by the pressure of this gas.

c) The pressure a gas would exert if it were alone in the vessel.


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Sample Question 24

Determine total pressure of a gas in a mixture of gases

When 0.050 mole of gas A are put in a totally empty flask X, it exerts a pressure of 22 mm Hg. When

0.20 mole of gas B are put in the same totally empty flask X, it exerts a pressure of 8 mm Hg. Bothquantities (of A and B) are now placed in an identical empty flask Y.

Choose ALL the correct answers about flask Y.

a)

The total pressure exerted by the gases is 0.25 mm Hg.b) The total pressure is 30 mm Hg.c) The pressure exerted by gas A, is only a part of 22 mm Hg (that is why they call it partial

pressure).d) The pressure exerted by gas B, is 85 mm Hg.

Sample Question 25

Know that the partial pressures ratio of gases equals their moles ratio

When 0.050 mole of gas A are put in a totally empty flask X, it exerts a pressure of 22 mm Hg. When

0.20 mole of gas B are put in the same totally empty flask X, it exerts a pressure of 8 mm Hg. Bothquantities (of A and B) are now placed in an identical empty flask Y.

Choose ALL the correct answers about flask Y.

a) ratio of moles of A to moles of B = ratio of partial pressure of A to partial pressure of B.

b) ratio of moles of A to total moles = ratio of partial pressure of A to total pressure.

c) ratio of moles of B to total moles = ratio of partial pressure of B to total pressure.

Sample Question 26

Mole fraction of gas A = ratio of partial pressure of A to total pressure

When 0.0050 mole of gas A are put in a totally empty flask X, it exerts a pressure of 22 mm

Hg. When 0.019 mole of gas B are put in the same totally empty flask X, it exerts a pressureof 85 mm Hg. Both quantities (of A and B) are now placed in an identical empty flask Y.

Choose ALL the correct answers about flask Y.a) Mole fraction of gas A = (0.0050/0.019).

b) Mole fraction of gas A = (0.0050/0.059).

c) Mole fraction of gas A = (22/85).

d) Mole fraction of gas A = (22/107).

Steps: = = + =

5.3 The ideal gas

5.3.1 Real gases


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5.3.2 Defining an ideal gas

Sample Question 27

Assumptions of the kinetic theory for an ideal gas

To explain the behavior of an ideal gas, we make all the following assumptions but one.Which one is not an assumption?

Particles of an ideal gas:

a)

are “point masses”. b) exert no forces on each other.

c) move in straight lines and undergo elastic collisions.

d) move in zigzag motion without colliding with other objects.e) have (an average) kinetic energy proportional to the absolute temperature.

Sample Question 28

How a real gas differs in behavior from an ideal gas

Which of the following statements describe a behavior of real gases which does not apply to an ideal

gas? Real gasesa) are made of tiny particles that move in straight lines unless they undergo collisions

b) liquefy at high pressures and low temperatures.

c) have particles that collide randomly with each other.d) have particles that collide at random, resulting in zigzag motion.

e) have particles that undergo elastic collisions.

f) only approximately obey P×V = constant.

5.3.3 The equation of an ideal gas

Pressure-temperature behavior of an ideal gas

Sample Question 29

Know the pressure-temperature behavior of an ideal gas

How are pressure and temperature related for an ideal gas?

a) P T

b) P ∝ tc) P ∝1/Td) P ∝1/te) P ∝ V


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Sample Question 30

Know when the pressure exerted by any gas becomes zeroWhen does the pressure exerted by any gas become zero?

a) at t = 0

b) at T = 0 (as the pressure of an ideal gas is directly proportional to the absolute

temperature)c) at STP

d)

at room temperature and pressuree) Only when the volume of the gas becomes practically zero

f) at -273.16°C

Sample Question 31

The volume-pressure behavior

Know the pressure-volume behavior of an ideal gasFor an ideal gas, how are the pressure and volume related?

a)

P

∝ V

b)

P 1/Vc) P ∝ T + td) P ∝ T +273e) P ∝ V+273

Pressure vs. number of moles

Sample Question 32

Know the pressure-moles behavior of an ideal gasFor an ideal gas, how are the pressure and number of moles related?

a)

P n b) P ∝ 1/nc) P.n ∝ constantd) P ∝ n +273e) P is independent of the number of moles.

The equation of state

Sample Question 33

Recognizing the equation of state of an ideal gas

Which of the following is the equation of state of an ideal gas?a) PV = b) PV =

c) PT = nRV

d) PV = nRT

e) = nRT


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Calculations based on the equation of state

Sample Question 34

Calculating the value of the universal gas constant

What data do we need in order to calculate the value of R?

a) P = 1.00 atm, V = 22.4 dm3, n = 1 mol, t = 100 °C

b)

P = 1.00 atm, V = 22.4 dm3, n = 1 mol, t = 273°Cc) P = 1.00 atm, V = 22.4 dm

3, n = 1 mol, T= 273K

d) P = 76 mmHg,V = 1.0 dm3, n = 1 mol, t = 0 °C

e) P = 1.00 atm, V = 22.4 dm3, n = 1 mol, T = 373K

Sample Question 35

Recognizing the value and units of the universal gas constant

Which part gives the value and units of R?

a) 0.082 dm3.atm. °C

-1mol

-1

b)

22.4/273 dm3

.atm.K -1

mol-1

c) 0.082 m

3.atm.K

-1mol

-1

d) 0.082 dm3.(mm Hg).K

-1mol

-1

e) 22.4/237 dm3.atm.K.mol

-1

f) 0.082 dm3.atm.K

-1mol

-1

5.4 Effusion of Gases

Sample Question 36

Know the meaning of effusion

Effusion meansa) Boiling.

b) Evaporating.

c) escaping through small opening(s).d) forming bubbles, like a fizzy drink in a bottle.

e) giving off a gas.

Sample Question 37

Effusion experiment shows that average molecular K.E = constant at constant temperature

Which generalization can be made from the effusion experiment?

a)

At the same temperature, the average molecular kinetic energy (½ mv 2

) for different gasesis the same: ½ mv

2 = constant.

b) At the same temperature, the molecules of all gases move at the same average speed.c) At the same pressure, the molecules of all gases have equal kinetic energy (on the average).


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Sample Question 38

Effusion of one gas at different temperature show that average molecular K.E ∝ TWhich generalization can be made from the effusion experiment?

a) Average molecular kinetic energy (½ mv 2) is proportional to the absolute temperature.

b) ½ Mv 2 T.

c) When the values of Mv2 are plotted against absolute temperature a straight line is obtained

that passes through the origin.

Sample Question 39

Know the meaning of diffusion

Define diffusion.

Diffusion is the term used to describe the mixing of gases.

It is the spread of a gas throughout space.

5.5 Experiment to determine the quantity of a gas

Sample Question 40

Know how to read the volume of a gas collected over water

When Mg reacted with HCl, H2 gas was produced and collected in a gas measuring tube. In order to readthe volume of the gas, we should

a) close the bottom of the tube with a thumb, take the tube out of water, then take the reading.

b) put a piece of white paper behind the level of the meniscus in the tube then take the reading.

c) move the tube up or down until the level of water inside and outside the tube is the same,

then take the reading.

d)

transfer the gas gently to a burette then take the reading.e) transfer the gas gently to a pipette then take the reading.

Sample Question 41

Know how to find the pressure of hydrogen gas collected over water

When Mg reacted with HCl, H2 gas was produced and collected in a gas measuring tube. In order to

determine the pressure of the hydrogen gas, we should:a) add the vapor pressure of water to the atmospheric pressure

b) subtract the vapor pressure of water from the atmospheric pressure

c)

just read the atmospheric pressured) first dry the hydrogen gas by passing it through a cotton towel

e) “pop” the hydrogen gas with pure oxygen and determine the mass of water formed

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Level L Chemistry BQ Chapter 5

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