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Chemistry 205 (UConcordia) > Chapter 10 > Flashcards

Chapter 10 Flashcards

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1
Q

how can we measure the atmospheric pressure?

A

The atmospheric pressure can be measured with a barometer.

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2
Q

what are the units of measurement that we can use to calculate the pressure?

A

millimeters of mercury (mm Hg)

standard atmospheres (atm)

the SI unit of pressure is the pascal (Pa)

bar

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3
Q

What Boyle observed when he studied the compressibility of gases ?

A

observed that the volume of a fixed amount of gas at a given temperature is inversely proportional to the pressure exerted by the gas

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4
Q

Mathematically what is the Boyle law?

A

p ∝ 1/V when n (amount of gas) and T (temperature) are constant

Where the symbol ∝ means “proportional to”.

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5
Q

When two quantities are proportional to each other, they can be equated if a proportionality constant, here called C_b. What is the mathematical formula?

A

P=C_b X 1/V or PV= C_b at constant n and T

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6
Q

P=C_b X 1/V. What this form of Boyle’s law express?

A

This form of Boyle’s law expresses the fact that the product of the pressure and volume of a gas sample is a constant at a given temperature, where the constant 𝐶𝑏 is determined by the amount of gas (in moles) and its temperature (in kelvins).

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7
Q

When Boyle’s law is useful?

A

Boyle’s law is useful when we want to know, for example, what happens to the volume of a given amount of gas when the pressure changes at a constant temperature

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8
Q

Boyle’s law

A bicycle pump has a volume of 1400 cm3. If a sample of air in the pump has a pressure of 730 mm Hg, what is the pressure when the volume is reduced to 170 cm3?

Example 10.2: Chemistry and chemical reactivity, 9th edition student book.

A

You can solve this problem by substituting data into a rearranged version of Boyle’s law.

p2= p1 (v1/v2)

p2= (730 mm Hg) X (1400 cm3 / 170 cm3)
= 6.0 X 10^3 mm Hg

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9
Q

What Jacques Charles discovered ?

A

Jacques Charles discovered that the volume of a fixed quantity of gas at constant pressure decreases with decreasing temperature

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10
Q

Who proposed the Kelvin scale?

A

William Thomson

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11
Q

When Kelvin temperatures are used with volume measurements what is the volume-temperature relationship ?

A

V=C_c × T

C_c is a proportionality constant (which depends on the amount of gas and its pressure).

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12
Q

What Charles’s law state?

A

Charles’s law states that if a given amount of gas is held at a constant pressure, its volume is directly proportional to the Kelvin temperature.

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13
Q

Charles’s law

A sample of CO2 in a gas-tight syringe has a volume of 25.0 mL at room temperature (20.0 C). What is the final volume of the gas if you hold the syringe in your hand to raise its temperature to 37.0 C (and hold the pressure constant)?

Example 10.3: Chemistry and chemical reactivity, 9th edition student book.

A

Step 1: convert celsius to kelvin

  1. 0 C = 20.0 + 273.2= 293.3 K
  2. 0 C = 37.0 + 273.2= 310.2 K

Step 2: Find the V2

V2= V2 X (T2/T1)
25.0 mL X (310 K / 293.2 K) = 26.4 mL

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14
Q

what is the general gas law?

A

The volume of a given amount of gas in inversely proportional to its pressure at constant temperature (Boyle’s law) and directly proportional to the Kelvin temperature at constant pressure (Charles’s law).

Combination of Boyle’s and Charles’s law

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15
Q

What is the equation of the general gas law?

A

(P1V1 / T1) = (P2V2 / T2)

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16
Q

When do we use the equation of the general gas law?

A

In applies specifically to situations in which the amount of gas does not change

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17
Q

General gas law

Helium-filled balloons are used to carry scientific instruments high into te atmosphere.
Suppose a balloon is launched when the temperature is 22.5 C and the barometric pressure if 754 mm Hg. If the balloon’s volume is 4.19 X 10^3 L, what will the volume be at a height of 20 miles, where the pressure is 76.0 mm hG and the temperature is -33.0 C?

Example 10.4: Chemistry and chemical reactivity, 9th edition student book.

A

Step 1: convert celsius to kelvin

  1. 5C= 22.5 + 273.2= 295.7
    - 33.0C= -33.0 + 273.2= 240.2
Step 2: put all your information clearly 
V1= 4.19 X 10^3 L
V2= ?
P1= 754 mm Hg
P2= 76 mm Hg
T1= 295.7 K
T1= 240.2 K

Step 3: Calculate

V2= V1 X (P1 / P2) X (T2 / T1)

4.19 X 10^3 L X (754 / 76) X (240.2 / 295.7)
=3.38 X 10^4 L

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18
Q

What is the Avogadro’s hypothesis?

A

He propose that equal volumes of gases under the same conditions of temperature and pressure have equal numbers of particles

Stated another way, the volume of a gas at a given temperature and pressure is directly proportional to the amount of gas in moles:
V ∝n at constant T and P

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19
Q

What is the ideal gas law formula?

A

PV=nRT

20
Q

What is the standard molar volume of gas?

A

Under conditions of standard temperature and pressure (STP)

  • a gas temperature of 0 Celsius or 273.15 Kelvin
  • a pressure of 1 atm

, 1 mole of an ideal gas occupies 22.414 L, a quantity called the standard molar volume.

21
Q

Ideal gas law

The nitrogen gas in an automobile air bag, with a volume of 65 L, exerts a pressure of 829 mm Hg at 25 celsius.

What amount of N2 gas (in moles) is in the air bag?

Example 10.6: Chemistry and chemical reactivity, 9th edition student book.

A

To use the ideal gas law with R having units of
L ∙ atm/ K ∙ mol, the pressure must be expressed in atmospheres and the temperature in kelvins.

Therefore step 1: convert the pressure and temperature to values with these units

P= 829 mm Hg X ( 1 atm / 760 mm Hg) = 1.09 atm
T=25 + 273= 298 K

Step 2: solve the equation
n= (PV) / (RT)= (1.09 atm X 65L) /
(0.082057 L∙atm/k∙mol X 298 k)
n=2.9 mol

22
Q

What is the formula to calculate the density of a gas?

A

d= PM / RT

23
Q

Density and molar mass

Calculate the density of CO2 at STP. Is CO2 more or less dense than air.

air density is 1.29 g/L at STP

Example 10.7: Chemistry and chemical reactivity, 9th edition student book.

A

d= PM / RT

d= (1 atm X 44 g/mol) / (0.082057 L∙atm/k∙mol X 273 k)

d= 1.96 g/L

CO2 is more dense than air

24
Q

Calculating the molar mass of a gas from P, V, and T data

You are trying to determine, by experiment, the formula of a gaseous compound you made to replace chlorofluorocarbons in air conditioners. You have determined the empirical formula is CHF2, but to determine the molecular formula you need to know the molar mass.

You find that a 0.100g sample of the compound exerts a pressure of 70.5 mm Hg in a 256 mL container at 22.3 celsius.

What is the molar mass of the compound?
What is its molecular formula?

Example 10.8: Chemistry and chemical reactivity, 9th edition student book.

A

–The density of the gas is the mass of the gas divided by the volume

D= 0.100g / 0.256 L = 0.391 g/L

–Use the density along with the values of pressure and temperature in equation d= PM / RT, and solve for the molar mass M

M= dRT / P=
(0.391 g/L X 0.082057 L∙atm/k∙mol X 295.5 K) / 0.0928 atm
= 102 g/ mol

–With this result, you can compare the experimentally determined molar mass with the mass of a mole of gas having the empirical formula CHF2.

experimental molar mass 102 g/mol
———————————— = ———————-
mass of 1 mol CHF2 51.0 g/formula unit

          = 2 formula units of CHF2 per mol

Therefore, the formula of the compound is C2H2F4.

–In the alternative approach, you use the ideal gas law to calculate the amount of gas, n.

n= (PV) / (RT)= ( 0.0928 atm X 0.256 L) / ( 0.082057 L∙atm/k∙mol X 295.5 K)

n= 9.80 X 10^-4 mol

–You know that 0.100 g of gas is equivalent to 9.80 X 10^-4 mol. Therefore,

molar mass = (0.100g) / ( 9.80 X 10^-4 mol)
molar mass = 102 g/mol

25
Q

Gas law and stoichiometry

Hydrogen gas is produced from the reaction of lithium and water. What mass of lithium is required to produce 23.5 L of H2 at 17.0 celsius and a pressure of 743 mm Hg?

The gas-producing reaction is

2Li (s) + 2 H2O (aq)= 2 LiOH (aq) + H2 (g)

Example 10.9: Chemistry and chemical reactivity, 9th edition student book.

A

Step 1: find the amount (mol) of gas produced

n=H2 produced (mol)= (PV) / (RT)
n= (0.978 atm X 25.5L) / (0.082057 L∙atm/k∙mol X 290.2 K)
n= 0.965 mol H2

Step 2: calculate the quantity of lithium that will produce 0.965 mol of H2 gas

Mass of Li = 0.965 mol H2 (2 mol Li / 1 mol H2) X (6.941 g / 1 mol Li) = 13.4 g Li

26
Q

what is the partial pressure?

A

the pressure of each gas in the mixture is called its partial pressure

27
Q

What is the Dalton’s law of partial pressures

A

John Dalton was the first to observe that the pressure of a mixture of ideal gases is the sum of the partial pressures of the different gases in the mixture

28
Q

How can we write the Dalton’s law mathematically?

A

P_total=P_1+P_2+P_3….

29
Q

How can we calculate the total pressure of a gaseous mixture?

A

Dalton’s law, the total pressure exerted by the mixture is the sum of the pressures exerted by each components

Complete formula
Ptotal=P_A + P_B + P_C =n_A(RT/V) + n_B(RT/V) +n_C(RT/V)

resumed formula
P_total=(n_A+n_B+n_C)(RT/V)
P_total=(n_total )(RT/V)

30
Q

How do we calculate the pressure exerted by each gas from the ideal gas law equation

A

P_A V=n_A RT
P_B V=n_B RT
P_C V=n_C RT

31
Q

for mixtures of gases, what is the mole fraction?

A

symbol: X

which is defined as the number of moles of a particular substance in a mixture divided by the total number of moles of all substance present

32
Q

which equation tells us that the pressure of a gas in a mixture of gases is the product of its mole fraction and the total pressure of the mixture

A

P_A=X_A P_total

33
Q

partial pressures of gases

Halothane, C2HBrClF3, is a nonflammable, nonexplosive, and nonirritating gas that was widely used as an inhalation anesthetic.

The total pressure of a mixture of 15.0g of halothane vapour and 23.5g of oxygen gas is 855mm Hg.

What is the partial pressure of each gas?

Example 10.10: Chemistry and chemical reactivity, 9th edition student book.

A

Step 1: calculate the mole fractions

Amount of C2HBrClF3= 15.0g ( 1 mol / 197.4g)= 0.0760 mol

Amount of O2 = 23.5g (1 mol / 32.00g) = 0.734 mol

total amount of gas =0.0760 mol C2HBrClF3 +0.734 mol O2
= 0.810 mol

mole fraction of C2HBrClF3 = 0.0760 / 0.810
= 0.0938

Because the sum of the mole fraction of halothane and of O2 must equal 1.00, this means that the mole fraction of oxygen is 0.906.

xhalothane + xoxygen= 1.00
0.0938 + xoxygen=1.00
xoxygen= 0.906

Step 2: Calculate partial pressures

partial pressure of halothane
= Phalothane = xhalothane X Ptotal
= 0.0938 X p total = 0.938 X 855 mm Hg
= 80.2 mm Hg

The total pressure of the mixture is the sum of the partial pressures of the gases in the mixture.

Phalothane + Poxygen = 855 mm Hg

and so

Poxygen= 855 mm Hg - Phalothane 
Poxygen= 855 mm Hg - 80.2 mm Hg

= 775 mm Hg

34
Q

what are the 3 main postulates in the kinetic-molecular theory of gases

A

gases consists of particles whose separation is much greater than the size of the particles themselves

the particles of a gas are in continual, random, and rapid motion. As they move, they collide with one another and with the walls of the container, but they do so in a way in which the total energy is unchanged.

The average kinetic energy of gas particles is proportional to the gas temperature. All gases, regardless of their molecular mass, have the same average kinetic energy at the same temperature.

35
Q

Does the molecules in a gas sample move at the same speed?

A

no, there is a distribution of speeds and the distribution depends on temperature.

36
Q

What is the equation of the kinetic energy of a single molecule of mass m (in kg) in a gas sample ?

A

KE=1/2 (mass X speed)^2

37
Q

How can you calculate the average kinetic energy of a mole of molecules in a gas sample?

A

KE=1/2 N_A (mass X speed)^2

N_A= Avogadro’s number

38
Q

Molecular speed

calculate the rms speed of oxygen molecules at 25 celsius.

Example 10.11: Chemistry and chemical reactivity, 9th edition student book.

A

step 1: find the molar mass of oxygen in Kg/mol

O2= 15.9994 X 2= 32 (rounded) g/mol

(32 g/mol) / (1000)= 0.032 kg/mol

Step 2: find the root-mean-square speed (rms speed)

√(u^2)= √ (3RT/M)
√(u^2)= √ ((3 (0.082057 L∙atm/k∙mol X 298) / 0.032 kg/mol))
√(u^2)= √232 000 J/Kg

Step 3: do the square root
√232 000 J/Kg= 482 J/Kg

Step 4: convert the answer in meter per second

1J= 1Kg ∙ m2/s2

(482 Kg ∙ m2) / (1 Kg ∙ s2)= 482 m/s

39
Q

How can we explain the gas law by the kinetic-molecular theory?

A

the starting place is to describe how pressure arise from collisions of gas molecules with the walls of the container holding the gas

gas pressure=(force of collisions) / (area)

The force exerted by the collisions depends on the number of collisions and the average force per collision

40
Q

When the temperature of a gas is increased, the average kinetic energy of the molecule___________. And what happens with the collisions?

A

When the temperature of a gas is increased, we know the average kinetic energy of the molecules increases

41
Q

What happens if we increase the number of molecules of a gas at a fixed temperature and volume ?

A

The number of collisions per second increases.

42
Q

What is the diffusion?

A

The mixing of molecules of two or more gases due to their random molecular motions is the result of diffusion

43
Q

What is the effusion?

A

Is the movement of gas through a tiny opening in a container into another container where the pressure is very low

44
Q

What Thomas Graham studied?

A

studied the effusion of gases

45
Q

What Thomas Graham have found in his study of the effusion of gases?

A

He have found that the rate of effusion of a gas – the amount of gas moving from one place to another in a given amount of time – is inversely proportional to the square root of its molar mass

46
Q

What is the formula to compare the rate of effusion of gases?

A

(rate of effusion of gas 1) ( (molar mass of gas 2) )
———————————– =√( —————————– )
(rate of effusion of gas 2) ( (molar mass of gas 1) )

47
Q

Using Graham’s law of effusion to calculate a molar mass

Tetrafluoroethylene, C2F4, effuses through a barrier at a rate of 4.6 X 10^-6 mol/h.

An unknown gas, consisting only of boron and hydrogen, effuses at the rate of 5.8 X 10^-6 mol/h under the same conditions.

What is the molar mass of the unknown gas?

Example 10.11: Chemistry and chemical reactivity, 9th edition student book.

A

(5. 8 X 10^-6 mol/h) ( (100.0 g / mol) )
- ————————- =√( —————– )
(4. 6 X 10^-6 mol/h) ( (M unknown ) )

      (    (100.0 g / mol)   ) 1.3= √(     -----------------       )
     (     (M unknown )    )

             100.0 g (1.3)^2= -----------------
          M unknown

          100.0 g 1.69 = -----------------
        M unknown

M unknown = 63.0 g/mol

Chemistry 205 (UConcordia) (11 decks)

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