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 6 of 20

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

Responses

Your email address will not be published.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

back-to-top
error:

Question 1

(a) State Charles law

Answer - Charles law states that the volume of a given mass of gas is directly proportional to its absolute temperature or temperature provided the pressure remains constant

(b) Draw a graphical representation of Charles Law

Answer -

(c) What is absolute zero temperature.

Answer - Absolute temperature is the temperature at which the molecules of a gas have no volume, no kinetic energy (no velocity), and no pressure.

State its value in 

   (i) Celsius temperature  

Answer - -2730C

   (ii) Kelvin temperature

Answer -   0K

Question 2

(a) A fixed mass of gas has a volume of 92cm3 at 30C. What will be its volume at 180C, if the pressure remains constant?

V1 = 92cm3

V2 = ?

T1 = 30C  ⇒ 276k

T2 = 180C  ⇒ 291k

Using the formula: 

:- \( \frac{V_1}{T_1}  = \frac{V_2}{T_2}  = \)

V2 = \(\frac{V_1 T_2}{T_1}  \\=\frac{92 \; \times \;  291}{276} \)

V2  = 97cm3

(b) Explain Charles Law in terms of Kinetic Theory

Answer - An increase in temperature gives rise to an increase in the average kinetic energy and the molecules of gas move more rapidly and collide with the walls of the container. At constant pressure, the volume of the container increase with an increase in temperature. Conversely, a decrease in temperature lowers the average kinetic energy of the gas molecules and also lowers the frequency of collision of gas molecules.

Question 3

(a) What volume would a gas occupy at s.t.p if at 430C and 720mmHg it occupies 214cm3

Answer - 

 T1 = 430C = 316k

T2 = 273k

V1 = 214cm3

V2 = ?

P1 = 720mmHg

P2 = 760mmHg

P1V1T2 = P2V2T1

V2 =  \( \frac{P_1V_1T_2}{P_2 T_1} = \frac{720 \; \times \; 214 \; \times \; 273}{760 \; \times \; 316} \)

= 175.15cm3

V2 = 175.15cm3

(b) A given mass of gas occupies 500cm3 at 300C and 6.5x10Nm-2. Calculate the volume of the gas at stp[Standard pressure = 1.01x105Nm-2]

Answer - 

 V1 = 500cm3

V2 = ?

T1 = 300C ⇒ 303k

T2 = 273k

P1 = 6.5 x1 0Nm-2

P2 = 1.01 x1 05

P1V1T2 = P2V2T1

V2 =  \( \frac{P_1V_1T_2}{P_2 T_1} = \frac{6.5 \; \times \; 10^1\; \times \; 500 \; \times \; 273}{1.01 \; \times \; 10^{-5}\; \times \; 303} \)

= 2899.23 x 10-4 or 2.9 x 10-2

Question 4

(a) State Dalton’s Law of partial pressure

Answer - In a mixture of gases that do not react chemically together, the total pressure exerted by the mixture is equal to the sum of the partial pressures of the individual gases present in the mixture.

(b) 210cm3 of Nitrogen at a pressure of 400mmHg and 180cm3 of Carbon(IV) Oxide at a pressure of 350mmHg were introduced into a 200cm3 vessel. What is the total pressure in the vessel?

Answer -

 

N2 CO2
Vol ⇒ 210cm3 Vol ⇒ 180cm3
Pressure ⇒ 400mmHg Pressure ⇒ 350mmHg

 

Volume of the vessel = 200cm3

 

:- \(\frac{Volume \; of \; the \;gas}{Volume \; of \; the \;vessel} \; \times \; \frac{Pressure \; of \;the\;gas}{1} \)

 

N2 CO2
\(\frac{210}{200} \; \times \; \scriptsize 400 \) \(\frac{180}{200} \; \times \; \scriptsize 350 \)
= 420mmHg = 315mmHg

According to Dalton’s law of partial pressure, 

Ptotal = Pa + Pb + Pc … + Pn

Ptotal = PN2 + PCO2

Ptotal = 735mmHg

Question 5

Two vessels contain separately 50cm3 of Nitrogen at 100mmHg pressure and 80cm3 of Oxygen at 200mmHg and they are connected by taps to a third vessel containing 30cm3 of Hydrogen at 500mmHg. If the taps are opened and the gases are allowed to mix until equilibrium is attained, determine

 

(a) The partial pressure of each of the three gases

 Answer - Total volume of the gases (N2 + O2 + H2) = 50 + 80 + 30 = 160cm3

N2 CO2 H2
Vol = 50cm3 Vol = 80cm3 Vol = 30cm3
Pressure = 100mmHg Pressure = 200mmHg Pressure = 500mmHg

 

Total volume of the gases = 160cm3

 

N2 CO2 H2
\(\frac{50}{160} \; \times \; \scriptsize 100 \) \(\frac{80}{160} \; \times \; \scriptsize 200 \) \(\frac{30}{160} \; \times \; \scriptsize 500 \)
= 31.25mmHg = 100mmHg  = 93.75mmHg

 

 (b) The total pressure of the gaseous mixture.

 Answer

Ptotal = Pa + Pb + Pc … + Pn

Ptotal = PN2 + PO2 + PH2

= (31.25 + 100 + 93.75)mmHg

= 225mmHg.

Question 1

(a) State Charles law

Answer - Charles law states that the volume of a given mass of gas is directly proportional to its absolute temperature or temperature provided the pressure remains constant

(b) Draw a graphical representation of Charles Law

Answer -

(c) What is absolute zero temperature.

Answer - Absolute temperature is the temperature at which the molecules of a gas have no volume, no kinetic energy (no velocity), and no pressure.

State its value in 

   (i) Celsius temperature  

Answer - -2730C

   (ii) Kelvin temperature

Answer -   0K

Question 2

(a) A fixed mass of gas has a volume of 92cm3 at 30C. What will be its volume at 180C, if the pressure remains constant?

V1 = 92cm3

V2 = ?

T1 = 30C  ⇒ 276k

T2 = 180C  ⇒ 291k

Using the formula: 

:- \( \frac{V_1}{T_1}  = \frac{V_2}{T_2}  = \)

V2 = \(\frac{V_1 T_2}{T_1}  \\=\frac{92 \; \times \;  291}{276} \)

V2  = 97cm3

(b) Explain Charles Law in terms of Kinetic Theory

Answer - An increase in temperature gives rise to an increase in the average kinetic energy and the molecules of gas move more rapidly and collide with the walls of the container. At constant pressure, the volume of the container increase with an increase in temperature. Conversely, a decrease in temperature lowers the average kinetic energy of the gas molecules and also lowers the frequency of collision of gas molecules.

Question 3

(a) What volume would a gas occupy at s.t.p if at 430C and 720mmHg it occupies 214cm3

Answer - 

 T1 = 430C = 316k

T2 = 273k

V1 = 214cm3

V2 = ?

P1 = 720mmHg

P2 = 760mmHg

P1V1T2 = P2V2T1

V2 =  \( \frac{P_1V_1T_2}{P_2 T_1} = \frac{720 \; \times \; 214 \; \times \; 273}{760 \; \times \; 316} \)

= 175.15cm3

V2 = 175.15cm3

(b) A given mass of gas occupies 500cm3 at 300C and 6.5x10Nm-2. Calculate the volume of the gas at stp[Standard pressure = 1.01x105Nm-2]

Answer - 

 V1 = 500cm3

V2 = ?

T1 = 300C ⇒ 303k

T2 = 273k

P1 = 6.5 x1 0Nm-2

P2 = 1.01 x1 05

P1V1T2 = P2V2T1

V2 =  \( \frac{P_1V_1T_2}{P_2 T_1} = \frac{6.5 \; \times \; 10^1\; \times \; 500 \; \times \; 273}{1.01 \; \times \; 10^{-5}\; \times \; 303} \)

= 2899.23 x 10-4 or 2.9 x 10-2

Question 4

(a) State Dalton’s Law of partial pressure

Answer - In a mixture of gases that do not react chemically together, the total pressure exerted by the mixture is equal to the sum of the partial pressures of the individual gases present in the mixture.

(b) 210cm3 of Nitrogen at a pressure of 400mmHg and 180cm3 of Carbon(IV) Oxide at a pressure of 350mmHg were introduced into a 200cm3 vessel. What is the total pressure in the vessel?

Answer -

 

N2 CO2
Vol ⇒ 210cm3 Vol ⇒ 180cm3
Pressure ⇒ 400mmHg Pressure ⇒ 350mmHg

 

Volume of the vessel = 200cm3

 

:- \(\frac{Volume \; of \; the \;gas}{Volume \; of \; the \;vessel} \; \times \; \frac{Pressure \; of \;the\;gas}{1} \)

 

N2 CO2
\(\frac{210}{200} \; \times \; \scriptsize 400 \) \(\frac{180}{200} \; \times \; \scriptsize 350 \)
= 420mmHg = 315mmHg

According to Dalton’s law of partial pressure, 

Ptotal = Pa + Pb + Pc … + Pn

Ptotal = PN2 + PCO2

Ptotal = 735mmHg

Question 5

Two vessels contain separately 50cm3 of Nitrogen at 100mmHg pressure and 80cm3 of Oxygen at 200mmHg and they are connected by taps to a third vessel containing 30cm3 of Hydrogen at 500mmHg. If the taps are opened and the gases are allowed to mix until equilibrium is attained, determine

 

(a) The partial pressure of each of the three gases

 Answer - Total volume of the gases (N2 + O2 + H2) = 50 + 80 + 30 = 160cm3

N2 CO2 H2
Vol = 50cm3 Vol = 80cm3 Vol = 30cm3
Pressure = 100mmHg Pressure = 200mmHg Pressure = 500mmHg

 

Total volume of the gases = 160cm3

 

N2 CO2 H2
\(\frac{50}{160} \; \times \; \scriptsize 100 \) \(\frac{80}{160} \; \times \; \scriptsize 200 \) \(\frac{30}{160} \; \times \; \scriptsize 500 \)
= 31.25mmHg = 100mmHg  = 93.75mmHg

 

 (b) The total pressure of the gaseous mixture.

 Answer

Ptotal = Pa + Pb + Pc … + Pn

Ptotal = PN2 + PO2 + PH2

= (31.25 + 100 + 93.75)mmHg

= 225mmHg.