12. Gas laws Questions and Answers

Gas laws Questions

1. A sample of unknown compound gas X is shown by analysis to contain Sulphur and Oxygen. The

gas requires 28.3 seconds to diffuse through a small aperture into a vacuum. An identical number

of oxygen molecules pass through the same aperture in 20seconds. Determine the molecular mass

of gas X (O= 16, S= 32)

2. (a) State Graham’s Law of diffusion

 (b) Gas V takes 10 seconds to diffuse through a distance of one fifth of a meter. Another

gas W takes the same time to diffuse through a distance of 10 cm. if the relative molecular

mass of gas V is 16.0; calculate the molecular mass of W

3. (a) State Charles’ Law

(b) The volume of a sample of nitrogen gas at a temperature of 291K and 1.0 x 10⁵ Pascals

was 3.5 x 10^-2m³. Calculate the temperature at which the volume of the gas would be

2.8 x 10^-2m³ at 1.0 x 10⁵pascals.

4. 60 cm³ of oxygen gas diffused through a porous partition in 50 seconds. How long would it take

60 cm³ of sulphur (IV) oxide gas to diffuse through the same partition under the sane conditions?

(S = 32.0, O = 16.0)

5. (a) State Graham’s law of diffusion

(b) 30cm³ of hydrogen chloride gas diffuses through a porous pot in 20seconds. How long

would it take 42cm³ of sulphur(IV) oxide gas to diffuse through the same pot under

the same conditions (H =1 Cl = 35.5 S = 32 O =16)

6. a) State Boyles law

b) Sketch a graph that represents Charles’ law

c) A gas occupied a volume of 250cm³ at -23ºC and 1 atmosphere. Determine its volume

at 127ºC when pressure is kept constant.

7. A factory produces Calcium Oxide from Calcium Carbonate as shown in the equation below:-

 CaCO_{3 (s)} CaO _(s) + CO_{2 (g)}

 (a) What volume of Carbon (IV) Oxide would be produced from 1000kg of Calcium

Carbonate at s.t.p (Ca = 40, C = 12, O = 16, Molar gas volume at s.t.p = 22.4dm³)

8. A fixed mass of gas occupies 200cm³ at a temperature of 23^oC and pressure of 740mmHg.

 Calculate the volume of the gas at -25^oC and 780mmHg pressure

9. Gas K diffuses through a porous material at a rate of 12cm³ s^-1 where as S diffuses through

the same material at a rate of 7.5cm³s^-1. Given that the molar mass of K is 16, calculate the

molar mass of S

10. (a) State Gay Lussac’s law

. 11. (a) What is the relationship between the rate of diffusion of a gas and its molecular mass?

 (b) A sample of Carbon (IV) Oxide takes 200 seconds to diffuse across a porous plug.

How long will it take the same amount of Carbon (II) Oxide to diffuse through the

same plug?(C=12, O=16)

12. Below are structures of particles. Use it to answer questions that follow. In each case only

electrons in the outermost energy level are shown

key

P = Proton

N = Neutron

X = Electron

(a) Identify the particle which is an anion

 (b) Choose a pair of isotopes. Give a reason

13. The figure below shows two gases P and Q diffusing from two opposite ends 18 seconds after

the experiment

(a) Which of the gases has a lighter density?

 (b) Given that the molecular mass of gas Q is 17, calculate the molecular mass of P

14. Identify the particles that facilitate the electric conductivity of the following substances

 (i) Sodium metal

 (ii) Sodium Chloride solution

(iii) Molten Lead Bromide

15. Gas B takes 110 seconds to diffuse through a porous pot, how long will it take for the

same amount of ammonia to diffuse under the same conditions of temperature and pressure?

(RMM of B = 34 RMM of ammonia = 17)

16. A gas occupies 5dm³ at a temperature of -27^oC and 1 atmosphere pressure. Calculate the

volume occupied by the gas at a pressure of 2 atmospheres and a temperature of 127^oC

17. A fixed mass of gas occupies 200 cm³ at a temperature of 23⁰c and a pressure of 740 mm Hg.

Calculate the volume of the gas at -25⁰c and 790 mm Hg pressure.

18. (a) State the Graham’s law

 (b) 100cm³ of Carbon (IV) oxide gas diffused through a porous partition in 30seconds.

How long would it take 150cm³ of Nitrogen (IV) oxide to diffuse through the same

partition under the same conditions? (C = 12.0, N = 14.0, O = 16.0)

Gas laws Answers

1. X: t₁= 28.3sec RMM = ?

Q₂: t₂= 20.0sec RMM=32

T  MM

T ₁
= X

T₂ 32

28.3
² = X

T₂ 32

X = 28.3² x 32

400

X = 64

2. (a) The rate of diffusion of a gas is inversely proportional to the square root of its density under the same conditions of temperature and pressure

(b) Rate of gas V= ¹/₅ x 100cm

10sec

= 2cm/sec

Rate of W = 10cm

10sec

= 1cm/sec

RV = MW

RW MV

2
² = MW

1 16

4 = MW = 4 x 16

1 16 1

MW = 64

3. (a) The volume of a fixed mass of a gas is directly proportional to its absolute temperature at

constant Pressure

 (b) Apply combined gas law; P₁V₁= P₂V₂

T₁ T₂

V₁ = 3.5 x10^-2 m³ V₂ = 2.8 x 10^-2m³

P₁ = 1.0 x 10⁵Pa P₂= 1.0 x 105Pa

T₁ = 291K T₂= ?

T₂ = P₂V₂T₁

P₁V₁

T₂ = 1.0 x 10⁵Pa x 2.8 x 10-²m³ x 291K

1.0 x 10⁵Pa x 3.5 x 10^-2m³

T₂ = 232.8k

4. TsO₂ = R.M.N.SO₂

½

 TO₂ R.M.MO₂

 SO₂ = 32 + (16 x 2) = 64 ½

 O₂ = (16 x 2) = 32 ½

 TsO₂ = 64
½ = 70.75 ½

50 32

5. a) The rate of diffusion of a fixed mass of a gas is inversely proportional to the square root of it

density at constant temperature and pressure

 b) RHCl = 30 cm ³ = 1.5 cm ³  see

20 se

RHCL = √MSO₂

 RSO₂ =√MHCL

 (1.5 ) ² √64

 RSO₂ = √ 36.5

 (RSO₂)² = 2.25 x 36.5

 64

 RSO_{2 =} √ 2.25 x 36.5

64

 1.133 cm/ sec

 1.133 cm³ ____________ 1 sec

 42 cm ³ = 42 x 1

1.133

= 37 sec

6. a) Boyles’ law For a fixed mass of a gas, volume is inversely promotional to pressure

at constant temperature

 b)

 c) P₁V₁ = P₂V₂

 ½ V₂ = P₁V₁ X T₂
 ½

T₁ T₂ T₁ P₂

250 X 273 – 23

273 + 127  ½

 = 156.5cm³

7. a) RFM of CaCO₃ = 40 + 12 + 48

= 100kg. ^√½

∴ 100 kg of CaCO₃
≡ 22.4dm³ of CO₂(g)

1000 kg ” ” ?

= 22.4 x 1000
^√1 = 224 dm^3
√½

 100

8. T₁ = 23+273 =296 T₂ = -25+273 = 248

V₁ = 200cm³ V₂ = ?

P_I = 740mmHg P₂ = 780mmHg

P₁V₁ = P₂V₂

T₁ T₂

740×200
√1=780x? √1

1. 248
   

x = 740 x 200 x 248

296 x 780

= 158.974 cm³√ 1 (penalize ½ mark for units)

9. Rk = Ms

Rs Mk

12 = x√ ½

7.2 16

X = 12² x 16√ ½

7.2²

= 44.464√

10. (a) When gases combine they do so in volume which bear a simple ratio to one another and to

the product if gaseous under standard temperature and pressure

11. a) Rate of diffusion is whereby proportional to molecular mass of a gas. √1

 b) TCO₂ = MCO₂

TCO MCO ^√½

⇒ 200 = 44 = 44 ^√½

T 28 28

⇒
200 = 11

T 7

⇒ T = 7

1. 11
   

⇒ T = 200.0.79772^√½ = 159.5 Seconds. ^√½

12. a) Y ^√1

 b) Z and W ^√1 have same atomic number but different mass number. ^√1

13. (a) Gas P

 (b) RQ = RMMP

RP RMMQ

18 = x

54 17

1² = x

3² 17

1 = x

9 17

9x = 17

x = ¹⁷/₉

x = 1.88

Q = It

= 5 x 386 = 1930C

(b) Pb²⁺_(l) + 2e Pb_(s) (½mk)

If 2 x 96500C = 207 ( ½ mk)

1930C = 1930 x 207 ( ½ mk)

2 x 96500

= 399510 (½mk)

193000C

= 2.07g (½mk)

14. i) Delocalized electrons

 ii) Mobile ions

 iii) Mobile ions

15. TNH₃ MNH₃

TB MB  ½

TNH₃  = 17

TB 34

TNH₃ = 17  ½

110 34

TNH3 = 110 X 17  ½ = 77.78 seconds  ½

34

16. P₁ V₁ = P₂ V₂

 T1 T₂

1×5 = 2 xV₂

246 400

 V2 = 400 x1 x 5

2 x246

= 4.065 dm³

17. a) V₁ = 200 cm³ V₂= ?

 T₁ = 296 K T₂ = 284K

 P₁ = 740 mmHg P₂ = 780 mm Hg

 P₁V_I = P₂V₂

T₁ T₂

 V₂ = P₁V₁T₂ = 740 mm Hg x 200cm³ x 248K

T₁P₂ 296K x 780 mm Hg

 = 158.97 cm³

 b) 60 l 1

18. a) Grahams law states

Under the same conditions of pressure and temperature, the rate of diffusion of a gas is

inversely proportional to the square root of its density

b) Time CO₂ = M_rCO₂

 Time NO₂ M_rNO₂

Where 100cm³ of CO₂ takes 30 seconds

 150cm3 of CO₂ takes ³⁰/₁₀₀ x 150

= 45 seconds√

45² = 0.975

TNO₂

45 = 44 ____ TNO₂ = 45

TNO₂ 46 0.978

TNO₂ = 46 sec

OR

RCO₂ = M_rNO₂

RNO₂ M_rCO₂

But RCO₂ = 100cm³ = 3.33 cm³ per sec

30 s

3.33 = 46

RNO₂ 44

= 1.0225

RNO₂ = 3.33

1.0225

= 3.26 cm ³ per second

Time for No = 150 cm³

3.26cm sec ^-1 = 46 secs