Back to Course

SS1: CHEMISTRY – 2ND TERM

0% Complete
0/0 Steps
  1. Kinetic Theory of Matter | Week 1
    3 Topics
    |
    2 Quizzes
  2. Kinetic Theory of Matter & Gas Laws 1 | Week 2
    3 Topics
    |
    2 Quizzes
  3. Gas Laws II | Week 3
    3 Topics
    |
    2 Quizzes
  4. Gas Laws III | Week 4
    3 Topics
    |
    2 Quizzes
  5. Gas Laws IV | Week 5
    3 Topics
    |
    2 Quizzes
  6. Mole Concept | Week 6
    5 Topics
    |
    2 Quizzes
  7. Acid, Bases and Salts I | Week 7
    7 Topics
    |
    2 Quizzes
  8. Acid, Bases and Salts II | Week 8
    6 Topics
    |
    2 Quizzes
  9. Acid, Bases and Salts III | Week 9
    3 Topics
    |
    2 Quizzes
  10. Acid, Bases and Salts IV | Week 10
    4 Topics
    |
    1 Quiz
  11. Acid, Bases and Salts V | Week 11
    5 Topics
    |
    1 Quiz
Quiz 10 of 20

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

Responses

Your email address will not be published. Required fields are marked *

back-to-top
error:

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 = \( \frac{1}{\sqrt{18}} = \scriptsize 0.24 \)