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1.0 dm3

of gasXat a pressure of 100 kPa and 2.0 dm3

of gas Yat a pressure of 200 kPa were

forced into a container of volume 0.50 dm3. Calculate

(a) the partial pressure of each gas in the mixture, and

(b) the total pressure of the mixture.

Assume the temperature remains constant throughout the whole process.

Solution 2.4

(a) For gasX:

Using P1V1 =P2V2(100 1.0) =P 0.50

PX = 200 kPa

For gas Y:

Using P1V1 =P2V2(200 2.0) =P 0.50

PY = 2800 kPa

(b) Total pressure = (200 + 800) kPa

= 1000 kPa

(a) Define an ideal gas.(b) Use the kinetic theory of gases to explain the following observations:

(i) The pressure of a fixed mass of gas increases with temperature at constant volume.

(ii) The volume of a fixed mass of gas decreases with increasing pressure at constant

temperature.

(iii) The volume of a fixed mass of gas increases with increasing temperature at constant

pressure.

Solution 2.5

(a) A gas that obeys the gas law:pV = nRTunder all conditions.

(b) (i) When temperature increases, the kinetic energy of the particles increases. This leads to

an increase in the rate of collision with the walls of the container, at the same time, thecollisions are more energetic.

(ii) Increasing pressure pushes the molecules closer to one another. This reduces the

amount of empty space between the particles, causing the volume to decrease.

(iii) When temperature increases, the pressure exerted by the gas will increase. To

maintain the original pressure, the volume occupied by the gas must increase so as to

reduce the rate of collision.

9CD Scripts

Question 2.4

Question 2.5

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The graph of pV

againstp for one mole of nitrogen gas at various temperatures is shown below.nRT

Explain the shape of the curves.

Solution 2.6

At pressure close to zero, nitrogen gas shows ideal behaviour.

At a particular temperature, deviation increases with increasing pressure. At moderate pressures,

presence of intermolecular attraction causes negative deviation. At high pressures, presence of

intermolecular repulsion makes the gas more difficult to compress, giving rise to positive

deviation. At a fixed pressure, deviation decreases as the temperature of the nitrogen gas

increases.

(a) State Daltons law of partial pressure.

(b) What do you understand by the partial pressure of a gas?

(c) 2.60 cm3

of argon at a pressure of 2.0 105

Pa, 12.5 dm3of ethane at a pressure of 1.2 10

5

Pa and 0.80 dm3

of carbon dioxide at 2.8 104

kPa were introduced into a 4.0 dm3vessel.

The temperature remains constant throughout the process.

(i) Calculate the partial pressure of each gas in the mixture.

(ii) What is the total pressure of the mixture?

(iii) If the carbon dioxide is removed from the mixture at constant temperature, calculate

the partial pressure of the argon and ethane in the container,

the total pressure of the mixture.

Solution 2.7

(a) Daltons law of partial pressure states that the total pressure exerted by a mixture of gases

that does not react with one another is equal to the sum of the partial pressures of the

constituent gases.

(b) (i) For argon:

(2.0 105) 2.60 = p 4.0

p(Ar) = 1.30 105

Pa

Question 2.6

Question 2.7

p/atm

PV

nRT

1 Ideal gas

0 100 200 300 400 500 600

100 K500 K

1200 K

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For ethane:

(1.2 105

) 12.5=p 4.0

p(Ethane) = 3.75 105

Pa

For CO2:

(2.8 104) 0.80 =p 4.0

p(CO2) = 5.60 103

Pa

(ii) Total pressure = (1.30 105) + (3.75 105) + (5.60 103) Pa

= 5.11 105

Pa

(iii) p(Ar) = 1.30 105

Pa

p(Ethane) = 3.75 105

Pa

Total pressure = (1.30 + 3.75) 105

Pa

= 5.05 105

Pa

Explain the following phenomena.

(a) Increasing pressure can sometimes cause a gas to liquefy.

(b) The volume occupied by one mole of ammonia at s.t.p. is less than 22.4 dm3.

(c) Gases quickly take up the shape of their container and always fill it.

(d) The noble gases become less ideal in their behaviour as one descends the group from helium

to xenon.

(e) A real gas is easier to compress at moderate pressure than at high pressure.

Solution 2.8

(a) Increasing pressure pushes the gas particles closer to one another. The attractive force is

then strong enough to hold the particles together causing them to condense.(b) Presence of intermolecular hydrogen bonding pulls the molecules closer.

(c) There are no intermolecular forces between gas particles. They are free to move around and

thus they will distribute themselves throughout the container.

(d) Going down the group, the size of the molecules gets bigger and the strength of the inter-

molecular forces between the molecules gets stronger, causing the deviation to be more

profound.

(e) At high pressures, intermolecular repulsion exists between the molecules. This makes the

gas more difficult to compress.

At moderate pressures, intermolecular attraction exists between the molecules, thus making

them easier to compress.

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Question 2.8