Quiz 10 of 20

# SS1: Chemistry 2nd Term Theory Questions – Gas Laws IV

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## Question 1

(a) state Graham’s Law of Diffusion

Answer -

Graham’s law of diffusion states that at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or relative molecular mass.

(b) 60cm3 of Hydrogen gas different through a Porous membrane in 10minutes. The same volume of gas G diffused through the same membrane in 37.4minutes. Determine the Relative molecular mass of G [H = 1 ]

Answer -

:- $$\frac{R_1}{R_2} = \frac{\sqrt{m_2}}{\sqrt{m_1}}$$

⇒$$\frac{RH_1}{R_G} = \frac{\sqrt{m_G}}{\sqrt{mH_2}}$$

At equal volume,

:- $$\frac{t_2}{t_1} = \frac{\sqrt{m_2}}{\sqrt{m_1}}$$

⇒ $$\frac{t_G}{tH_2} = \frac{\sqrt{m_G}}{\sqrt{H_2}}$$

:- $$\frac{37.4}{10} = \frac{\sqrt{m_G}}{\sqrt{2}}$$

Square both sides,

:- $$\frac{(37.4)^2}{(10)^2} = \frac{m_G}{2}$$

:- $$\frac{1,398.78}{100} = \frac{m_G}{2}$$

Cross multiply,

2,797.52 = 100MG

MG  = $$\frac {2,797.52}{100}$$

MG = 27.98

## Question 2

(a) Give three conditions under which Real gas behave Ideally

Answer -

1. At low pressure
2. Under laboratory condition
3. At high temperature

(b) How many moles of Oxygen gas are there in 350cm3 of the gas at 00C and 380mmHg pressure. (O2 =32, R =0.082 atm dm-3k-1mol-1)

Answer -

V = 350cm3 ⇒ 0.35dm3

P = 380mmHg

T = 273k

n = ?

R = 0.082 atm dm-3k-1mol-1

n = $$\frac{PV}{RT}$$

n = $$\frac{380 \; \times \; 0.35}{0.082 \; \times \; 273}$$

n = $$\frac{133}{22.39}$$

n = 5.95moles

## Question 3

(a) Show that for an Ideal gas PV=nRT, and define all the terms in the expression.

Answer -

:- $$\scriptsize P \propto T$$

:- $$\scriptsize P \propto \normalsize \frac{1}{V}$$ .............(2)

:- $$\scriptsize P \propto n$$ .............(2)

:- $$\scriptsize P \propto n \normalsize \frac{T}{V}$$

:- $$\scriptsize PV \propto nT$$

:- $$\scriptsize PV \propto KnT$$  K = constant of proportionality

PV = nRT ; where K = R

PV = nRT

(b) Calculate the pressure at 270C OF 16.0g of Oxygen gas occupying 2.50dm3

Answer -

PV = nRT

P = ?

V = 2.50dm3

n = $$\frac{m}{M} = \frac{16}{32} = \scriptsize 0.5\: moles$$

m = 16g

M = 32

R = 8.31Jk-1mol-1

T = 300k

P = $$\frac{nRT}{V}$$

P = $$\frac{0.5 \; \times \; 8.31 \; \times 300}{3.50}$$

P = $$\frac{1246.5}{2.50}$$

= 498.6 PA

## Question 5

Arrange the following gases in ascending order of their rates of diffusion

[S=32, Cl=35.5, H=1, O=16, C=12]

(a) Chlorine, Hydrogen, Oxygen, Carbon(IV) Oxide

Answer -

:- $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{m}}$$

(Cl) Chlorine = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{70}} \\ = \scriptsize 0.12$$

(H2) Hydrogen = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{2}} \\ = \scriptsize 0.71$$

(O2) Hydrogen = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{32}} \\ = \scriptsize 0.18$$

(CO2) Carbon (IV) Oxide = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{44}} \\ = \scriptsize 0.15$$

(b) Sulphur(IV) Oxide, Ammonia, Nitrogen

Answer -

SO2 = 64;

NH3 = 17;

N2 = 28

SO2 = $$\frac{1}{\sqrt{64}} = \scriptsize 0.13$$

NH3  = $$\frac{1}{\sqrt{17}} = \scriptsize 0.24$$

N2 = $$\frac{1}{\sqrt{28}} = \scriptsize 0.19$$

(c) Methane, Hydrogen Sulphide, Carbon(IV) Oxide, Hydrogen

Answer -

CH3 = 15

H2S = 34

CO2 = 44

H2 = 2

CH3 = $$\frac{1}{\sqrt{15}} = \scriptsize 0.25$$

H2S = $$\frac{1}{\sqrt{34}} = \scriptsize 0.17$$

CO2 = $$\frac{1}{\sqrt{44}} = \scriptsize 0.15$$

H2 = $$\frac{1}{\sqrt{2}} = \scriptsize 0.71$$

CO2 < H2S < CH3 < H2

(d) Ethene, ethane ethyne, Water vapour

Answer -

C2H4 =28;

C2H6 = 30;

C2H2 = 26;

H2O = 18

C2H4 = $$\frac{1}{\sqrt{28}} = \scriptsize 0.19$$

C2H6 = $$\frac{1}{\sqrt{30}} = \scriptsize 0.18$$

C2H2 = $$\frac{1}{\sqrt{26}} = \scriptsize 0.20$$

H2O = $$\frac{1}{\sqrt{18}} = \scriptsize 0.24$$

## Question 1

(a) state Graham’s Law of Diffusion

Answer -

Graham’s law of diffusion states that at constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or relative molecular mass.

(b) 60cm3 of Hydrogen gas different through a Porous membrane in 10minutes. The same volume of gas G diffused through the same membrane in 37.4minutes. Determine the Relative molecular mass of G [H = 1 ]

Answer -

:- $$\frac{R_1}{R_2} = \frac{\sqrt{m_2}}{\sqrt{m_1}}$$

⇒$$\frac{RH_1}{R_G} = \frac{\sqrt{m_G}}{\sqrt{mH_2}}$$

At equal volume,

:- $$\frac{t_2}{t_1} = \frac{\sqrt{m_2}}{\sqrt{m_1}}$$

⇒ $$\frac{t_G}{tH_2} = \frac{\sqrt{m_G}}{\sqrt{H_2}}$$

:- $$\frac{37.4}{10} = \frac{\sqrt{m_G}}{\sqrt{2}}$$

Square both sides,

:- $$\frac{(37.4)^2}{(10)^2} = \frac{m_G}{2}$$

:- $$\frac{1,398.78}{100} = \frac{m_G}{2}$$

Cross multiply,

2,797.52 = 100MG

MG  = $$\frac {2,797.52}{100}$$

MG = 27.98

## Question 2

(a) Give three conditions under which Real gas behave Ideally

Answer -

1. At low pressure
2. Under laboratory condition
3. At high temperature

(b) How many moles of Oxygen gas are there in 350cm3 of the gas at 00C and 380mmHg pressure. (O2 =32, R =0.082 atm dm-3k-1mol-1)

Answer -

V = 350cm3 ⇒ 0.35dm3

P = 380mmHg

T = 273k

n = ?

R = 0.082 atm dm-3k-1mol-1

n = $$\frac{PV}{RT}$$

n = $$\frac{380 \; \times \; 0.35}{0.082 \; \times \; 273}$$

n = $$\frac{133}{22.39}$$

n = 5.95moles

## Question 3

(a) Show that for an Ideal gas PV=nRT, and define all the terms in the expression.

Answer -

:- $$\scriptsize P \propto T$$

:- $$\scriptsize P \propto \normalsize \frac{1}{V}$$ .............(2)

:- $$\scriptsize P \propto n$$ .............(2)

:- $$\scriptsize P \propto n \normalsize \frac{T}{V}$$

:- $$\scriptsize PV \propto nT$$

:- $$\scriptsize PV \propto KnT$$  K = constant of proportionality

PV = nRT ; where K = R

PV = nRT

(b) Calculate the pressure at 270C OF 16.0g of Oxygen gas occupying 2.50dm3

Answer -

PV = nRT

P = ?

V = 2.50dm3

n = $$\frac{m}{M} = \frac{16}{32} = \scriptsize 0.5\: moles$$

m = 16g

M = 32

R = 8.31Jk-1mol-1

T = 300k

P = $$\frac{nRT}{V}$$

P = $$\frac{0.5 \; \times \; 8.31 \; \times 300}{3.50}$$

P = $$\frac{1246.5}{2.50}$$

= 498.6 PA

## Question 5

Arrange the following gases in ascending order of their rates of diffusion

[S=32, Cl=35.5, H=1, O=16, C=12]

(a) Chlorine, Hydrogen, Oxygen, Carbon(IV) Oxide

Answer -

:- $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{m}}$$

(Cl) Chlorine = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{70}} \\ = \scriptsize 0.12$$

(H2) Hydrogen = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{2}} \\ = \scriptsize 0.71$$

(O2) Hydrogen = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{32}} \\ = \scriptsize 0.18$$

(CO2) Carbon (IV) Oxide = $$\scriptsize R \propto \normalsize \frac{1}{\sqrt{44}} \\ = \scriptsize 0.15$$

(b) Sulphur(IV) Oxide, Ammonia, Nitrogen

Answer -

SO2 = 64;

NH3 = 17;

N2 = 28

SO2 = $$\frac{1}{\sqrt{64}} = \scriptsize 0.13$$

NH3  = $$\frac{1}{\sqrt{17}} = \scriptsize 0.24$$

N2 = $$\frac{1}{\sqrt{28}} = \scriptsize 0.19$$

(c) Methane, Hydrogen Sulphide, Carbon(IV) Oxide, Hydrogen

Answer -

CH3 = 15

H2S = 34

CO2 = 44

H2 = 2

CH3 = $$\frac{1}{\sqrt{15}} = \scriptsize 0.25$$

H2S = $$\frac{1}{\sqrt{34}} = \scriptsize 0.17$$

CO2 = $$\frac{1}{\sqrt{44}} = \scriptsize 0.15$$

H2 = $$\frac{1}{\sqrt{2}} = \scriptsize 0.71$$

CO2 < H2S < CH3 < H2

(d) Ethene, ethane ethyne, Water vapour

Answer -

C2H4 =28;

C2H6 = 30;

C2H2 = 26;

H2O = 18

C2H4 = $$\frac{1}{\sqrt{28}} = \scriptsize 0.19$$

C2H6 = $$\frac{1}{\sqrt{30}} = \scriptsize 0.18$$

C2H2 = $$\frac{1}{\sqrt{26}} = \scriptsize 0.20$$

H2O = $$\frac{1}{\sqrt{18}} = \scriptsize 0.24$$