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Class 9 SELINA Solutions Chemistry Chapter 7 - Study of Gas Laws

Study of Gas Laws Exercise Ex. 7

Solution Num 1

V1 = 500 dm3

P1 = 1 bar

T1 = 273 K

V2 = 500 dm3

T2 = 273 K

P2= ?

Solution 1

Gas is a state of matter in which interparticle attraction is weak and interparticle space is so large that the particles become completely free to move randomly in the entire available space.

Solution 2

  1. Gases are made of tiny particles which move in all possible directions at all possible speeds. The size of molecules is small as compared to the volume of the occupied gas.
  2. There is no force of attraction between gas particles or between the particles and the walls of the container. So, the particles are free to move in the entire space available to them.
  3. The moving particles of gas collide with each other and with the walls of the container. Because of these collisions, gas molecules exert pressure. Gases exert the same pressure in all directions.
  4. There is large interparticle space between gas molecules, and this accounts for high compressibility of gases.
  5. Volume of a gas increases with a decrease in pressure and increase in temperature.
  6. Gases have low density as they have large intermolecular spaces between their molecules.
  7. Gases have a natural tendency to mix with one and other because of large intermolecular spaces. So, two gases when mixed form a homogeneous gaseous mixture.
  8. The intermolecular space of a gas is reduced because of cooling. Molecules come closer resulting in liquefaction of the gas. 

Solution 3

The phenomenon is diffusion. In air, gas molecules diffuse to mix thoroughly. Hence, we can smell hydrogen sulphide gas from a distance in the laboratory.

Solution 4

Diffusion is the process of gradual mixing of two substances, kept in contact, by molecular motion.

Example:

If a jar of chlorine is opened in a large room, the odorous gas mixes with air and spreads to every part of the room. Although chlorine is heavier than air, it does not remain at the floor level but spreads throughout the room.

Solution 5

Temperature affects the kinetic energy of molecules. So, molecular motion is directly proportional to temperature.

Solution 6

  1. Three variables for gas laws: Volume (V), Pressure (P), Temperature (T)
  2. SI units of these variables:

For volume: Cubic metre (m3)

For pressure: Pascal (Pa)

For temperature: Kelvin (K) 

Solution 7

  1. Boyle's law: At constant temperature, the volume of a definite mass of any gas is inversely proportional to the pressure of the gas.  

Or

Temperature remaining constant, the product of the volume and pressure of the given mass of a dry gas is constant

i. Mathematical representation: 

According to Boyle's Law,

where K is the constant of proportionality.

If V' and P' are some other volume and pressure of the gas at the same temperature, then

 ii. Graphical representation of Boyle's Law:

1.  : Variation in volume (V) plotted against (1/P) at a constant temperature: A straight line passing through the origin is obtained.

 

 

2. V vs P: Variation in volume (V) plotted against pressure (P) at a constant temperature: A hyperbolic curve in the first quadrant is obtained.

 

  

3. PV vs P: Variation in PV plotted against pressure (P) at a constant temperature: A straight line parallel to the X-axis is obtained.

 

 iii. Significance of Boyle's law:

According to Boyle's law, on increasing pressure, volume decreases. The gas becomes denser. Thus, at constant temperature, the density of a gas is directly proportional to the pressure.

At higher altitude, atmospheric pressure is low so air is less dense. As a result, lesser oxygen is available for breathing. This is the reason mountaineers carry oxygen cylinders. 

Solution 8

Boyle's law on the basis of the kinetic theory of matter:

  • According to the kinetic theory of matter, the number of particles present in a given mass and the average kinetic energy is constant.
  • If the volume of the given mass of a gas is reduced to half of its original volume, then the same number of particles will have half the space to move.
  • As a result, the number of molecules striking the unit area of the walls of the container at a given time will double and the pressure will also double.
  • Alternatively, if the volume of a given mass of a gas is doubled at constant temperature, the same number of molecules will have double the space to move.
  • Thus, the number of molecules striking the unit area of the walls of a container at a given time will become one-half of the original value.
  • Thus, pressure will also get reduced to half of the original pressure. Hence, it is seen that if the pressure increases, the volume of a gas decreases at constant temperature, which is Boyle's law. 

 

Solution 9

  1. Pressure will double.
  2. Pressure remains the same. 

Solution 10

Charles's Law

At constant pressure, the volume of a given mass of a dry gas increases or decreases by 1/273rd of its original volume at 0°C for each degree centigrade rise or fall in temperature.

V T (at constant pressure)

 

At temperature T1 (K) and volume V1 (cm3):

  ...(i)

At temperature T2 (K) and volume V2 (cm3):

  ….(ii)

From (i) and (ii),

For Temperature = Conversion from Celsius to Kelvin

1 K = °C + 273

Example:

20°C = 20 + 273 = 293 K

 

Graphical representation of Charles's law

T vs V: The relationship between the volume and the temperature of a gas can be plotted on a graph. A straight line is obtained.

  

Graphical representation of Charles's law

 

Significance of Charles's Law: The volume of a given mass of a gas is directly proportional to its temperature; hence, the density decreases with temperature. This is the reason that

(a) Hot air is filled in balloons used for meteorological purposes. (b) Cable wires contract in winters and expand in summers. 

Solution 11

Charles's law on the basis of the kinetic theory of matter:

According to the kinetic theory of matter, the average kinetic energy of gas molecules is directly proportional to the absolute temperature. Thus, when the temperature of a gas is increased, the molecules would move faster and the molecules will strike the unit area of the walls of the container more frequently and vigorously. If the pressure is kept constant, the volume increases proportionately. Hence, at constant pressure, the volume of a given mass of a gas is directly proportional to the temperature (Charles's law). 

Solution 12

Absolute zero

The temperature -273°C is called absolute zero.

Absolute or Kelvin scale of temperature

The temperature scale with its zero at -273°C and each degree equal to one degree on the Celsius scale is called Kelvin or the absolute scale of temperature.

Conversion of temperature from Celsius scale to Kelvin scale and vice versa

The value on the Celsius scale can be converted to the Kelvin scale by adding 273 to it.

Example:

20°C = 20 + 273 = 293 K 

Solution 13

  1. The behaviour of gases shows that it is not possible to have temperature below 273.15°C. This act has led to the formulation of another scale known as the Kelvin scale. The real advantage of the Kelvin scale is that it makes the application and the use of gas laws simple. Even more significantly, all values on the Kelvin scale are positive.
  2. The boiling point of water on the Kelvin scale is 373 K. Now, K = °C + 273 and °C = K - 273.

The Kelvin scale can be converted to the degree Celsius scale by subtracting 273 So, the boiling point of water on the centigrade scale is 373 K - 273 = 100°C. 

Solution 14

  1. Standard or Normal Temperature and Pressure (STP or NTP)

The pressure of the atmosphere which is equal to 76 cm or 760 mm of mercury and the temperature is 0°C or 273 K is called STP or NTP. The full form of STP is Standard Tempe rature and Pressure, while that of NTP is Normal Temperature and Pressure.

Value: The standard values chosen are 0°C or 273 K for temperature and 1 atmospheric unit (atm) or 760 mm of mercury for pressure.

The standard values chosen are 0°C or 273 K for temperature and 1 atmospheric unit (atm) or 760 mm of mercury for pressure.

Standard temperature = 0°C = 273 K

Standard pressure = 760 mmHg

= 76 cm of Hg

= 1 atmospheric pressure (atm) 

 

  1. The volume of a given mass of dry enclosed gas depends on the pressure of the gas and the temperature of the gas in Kelvin, so to express the volume of the gases, we compare these to STP. 

Solution 15

  1.  
    1. °C = O°C
    2. K = 273 K
  1.  
    1. 1 atm
    2. 760 mmHg
    3. 76 cmHg
    4. 1 torr = 133.32 Pascal 

Solution 16

Temperature on the Kelvin scale (K)

= 273 + temperature on the Celsius scale

Or K = 273 + °C

 

(i) 273°C in Kelvin

    t°C = t K - 273

    273°C = t K - 273

    t K = 273 + 273 = 546 K

     273°C = 546 K

 

(ii) 293 K in °C

    t°C = 293 - 273

    t°C = 20°C

     293 K = 20°C 

Solution 17

  1. Charles's law
  2. Boyle's law 

Solution 18

  1. The advantage of the Kelvin scale is that it makes the application and use of gas laws simple. Of more significance is that all values on the scale are positive, removing the problem of negative (-) values on the Celsius scale.
  2. The mass of a gas per unit volume is small because of the large intermolecular spaces between the molecules. Therefore, gases have low density. In solids and liquids, the mass is higher and intermolecular spaces are negligible.
  3. At a given temperature, the number of molecules of a gas striking against the walls of the container per unit time per unit area is the same. Thus, gases exert the same pressure in all directions.
  4. Because the volume of a gas changes remarkably with a change in temperature and pressure, it becomes necessary to choose a standard value of temperature pressure.
  5. According to Boyle's law, the volume of a given mass of a dry gas is inversely proportional to its pressure at constant temperature.

 

When a balloon is inflated, the pressure inside the balloon decreases, and according to Boyle's law, the volume of the gas should increase. But this does not happen. On inflation of a balloon along with reduction of pressure of air inside the balloon, the volume of air also decreases, violating Boyle's law.

  1. Atmospheric pressure is low at high altitudes. The volume of air increases and air becomes less dense because volume is inversely proportional to density. Hence, lesser volume of oxygen is available for breathing. Thus, mountaineers have to carry oxygen cylinders with them.
  2. Interparticle attraction is weak and interparticle space is large in gases because the particles are completely free to move randomly in the entire available space and takes the shape of the vessel in which the gas is kept. 

Solution 19

  1. The temperature scale with its zero at -273°C and where each degree is equal to the degree on the Celsius scale is called the absolute scale of temperature.
  2. The temperature -273°C is called absolute zero. Theoretically, this is the lowest temperature which can never be reached. All molecular motion ceases at this temperature.
  3. The temperature -273°C is called absolute zero.

Solution 20

Gases such as nitrogen and hydrogen are collected over water as shown in the diagram. When the gas is collected over water, the gas is moist and contains water vapour. The total pressure exerted by this moist gas is equal to the sum of the partial pressures of the dry gas and the pressure exerted by water vapour. The partial pressure of water vapour is also known as aqueous tension.

  

Collection of gas over water 

 

  Ptotal = Pgas + Pwater vapour

 Pgas = Ptotal- Pwater vapour

Actual pressure of gas = Total pressure - Aqueous tension 

Solution 21

  1. Volume of gas is zero.
  2. Absolute temperature is 7 + 273 = 280 K.
  3. The gas equation is

  1. Ice point = 0 + 273 = 273 K 
  2. Standard temperature is taken as 273 K or O°C.  

Standard pressure is taken as 1 atmosphere (atm) or 760 mmHg. 

Solution 22

  1. (iii) Straight line parallel to the X-axis.
  2. (ii) 27°C = 27 + 273 = 300 K
  3. (iv) Charles
  4. (ii) 1/2 times 

Solution 23

 

Column A

Column B

(a) cm3 

Volume

(b) Kelvin 

Temperature

(c) Torr 

 Pressure

(d) Boyle's law 

PV = P1 V1

 

(e) Charles's law 

  

 

 

Solution 24

  1. Volume of a gas is directly proportional to the pressure at constant temperature.
  2. Volume of a fixed mass of a gas is inversely proportional to the temperature, the pressure remaining constant.
  3. -273°C is equal to zero Kelvin.
  4. Standard temperature is 0°C.
  5. The boiling point of water is -373 K. 

Solution 25

  1. Absolute temperature
  2. Absolute zero
  3. Volume
  4. 273 

Solution Num 2

V = 2 litres

P = 760 mm

V1 = 4000 m3 [1 dm3 = 4 litres]

P1= ?

  

Solution Num 3

  1.  

 

P/atm

V/dm3

1/V

PV

0.2

112

0.009

22.4

0.25

89.2

0.011

22.4

0.4

56.25

0.018

22.4

0.6

37.4

0.027

22.4

0.8

28.1

0.036

22.4

1

22.4

0.045

22.4

 

 

 i. P vs. V:

 

  

At constant temperature, P is inversely proportional to V. Thus, the plot of V versus P will be a rectangular hyperbola.

 

 ii. P vs. 1/V:

 

  

 

According to Boyle's law, at constant temperature, pressure of a fixed amount of gas varies inversely to its volume. The graph of pressure verses 1/V shows a positive slope.

 

 iii. PV vs. P:

 

  

According to Boyle's law, the product of pressure and volume is constant at constant temperature. The graph of PV versus P is constant which indicates that the given gas obeys Boyle's law.

 

  1. The correct measurements of the volume are given below:

 

P/atm

V/dm3

0.2

112

0.25

89.6

0.4

56

0.6

37.33

0.8

28

1

22.4

 

Solution Num 4

Given: 

V = 800 cm3

P = 650 m

P1= ?

V1 = reduced volume = 40% of 800

 =

Net V1 = 800 - 320 = 480 cm3  

T = T1

Using the gas equation,

  

Since T = T1

Solution Num 5

 

Solution Num 6

V = 20 litre  

P = 29 atm

P1 = 1.25 atm 

V1 =? 

T = T1

  

Solution Num 7

Initial volume = V1 = 561 dm3

Final volume = V2 = 748 dm3

Difference in volume = 748 - 561 = 187 dm3

As the temperature is constant,

Decrease in pressure percentage =

  

Solution Num 8

V = 88 cm3 

P = 770 mm

P1 = 880 mm 

V1= ? 

T = T1

Volume diminishes = 88 - 77 = 11 cm3  

Solution Num 9

Let volume = 100 ml

T = 240 K

Volume increased = x ml

New volume = 100 + x ml

T1 = 400 K

  

Solution Num 10

(a) V1 = 0.4 L 

     V2 = 0.4 × 2L

     T1 = 17°C (17 + 273) = 290 K

     T2= ?

       

 

(b) V1 = 0.4 L 

     V2 = 0.2 L 

     T1 = 17°C (17 + 273) = 290 K

     T2= ?

      

Solution Num 11

V1 = 143 cm3

T1 = 27oC = 27+273 = 300K

P1 = 700 mm Hg

T2 = 300K

P2= 280 mm Hg

V2= ?

  

  2 = 357.5 cm3 

Solution Num 12

V = 500 cm3 

Normal temperature, t = 0°C = 0 + 273 K

V1 = Reduced volume + 20% of 500 cm3

Net, V1  = 500 - 100 = 400 cm3

T1= ?

P = P1

  

Solution Num 13

V1 = X

P1 = 1 atm

V2= ?

T2 = 3 T1

P2 = 2 atm

Solution Num 14

V = 30 cm3 

P = 740 mm

T = 288 K

P1 = 760 mm

V1= ? 

T1 = 273 K

Solution Num 15

V1 = V1

P1 = 760 atm

T1 = 273 K

V2 = 2V1

T2 =?

Solution Num 16

V = 100 cm3 

P = 760 mm

T = 273 K

V1= ? 

  

Solution Num 17

Let the original volume (V) = 1 and

the original pressure (P) = 1 and

the temperature given (T) = -15°C  = -15 + 273 = 258 K

V1 or new volume after heating = original volume + 50% of original volume

P1 or decreased pressure = 60%

 

T1 = to be calculated

  

Solution Num 18

V = 2 litres

P = 100 Pa

T = 300 K

  

T1 = 75 - 273 = -198°C

Solution Num 19

V1 = 2500 cm3

P1 = 1 atm = 760 mm

T1 = 273 K

V2= ?

T2 = 273 K

Solution Num 20

V1 = 714.29 cm3

P1 = P2 = P

T1 = 273 K

V2 = 2500 cm3

T2= ?

  

 

Solution Num 21

  1. V1 = V2 = V

       P1 = 100 cmHg

       T1 = 273 K

       P2 = 10 cmHg

       T2= ?

        

  1. V1 = V2 = V

       P1 = 100 cmHg

       P2= ?

       T1 = 273 K

       T2 = 373 K

        

Solution Num 22

Capacity of the cylinder V = 10000 litres

P = 800 mm

T = -3°C = -3 + 273 = 270 K P1 = 400 mmHg

T1 = 0°C = 0 + 273 = 273 K

V1= ?

  

Solution Num 23

V1 = 22.4 litres

P1 = 1 atm

T1 = 273 K

V2 =?

T2 = 300 K

P2 = 4 atm

Solution Num 24

V = 50 cm3 

P = 750 - 14 = 736 mm

T = 290 K

P1 = 760 mm

V1= ? 

T1 = 273 K

  

Solution Num 25

  1. V = 1.2 litres

       P = 748 mmHg

       T = 298 K

       P1 = 760 mmHg

       T1 = 273 K

       V1= ?

        

  1. V = 1.25 litres

      P = 760 mmHg

      T = 273 K

      P1 = 760 mmHg

      T1 = 273 K

      V1= ?

       

     1.25 litres O2 will have greater volume than 1.2 litres N2. 

Solution Num 26

Pressure due to dry air,

P = 750 - 12 = 738 mm

V = 28 cm3

T = 14°C = 14 + 273 = 287 K

P1 = 760 mmHg

V1= ?

T1 = 0°C = 273 K

Using gas equation,

  

Solution Num 27

P = 14.9 atm

V = 28 cm3

T = ?

P1 = 12 atm

V = V1

T1 = 300 K

Using gas equation,

Solution Num 28

Step 1:

V1 = 20 litres

P1 = 700 mm

T1 = 300 K

V2= ?

T2 = 273 K

P2 = 760 mm

 

Step 2:

22.4 litres of the gas at STP weighs = 70 g

16.76 litres of the gas has weight at STP =

 

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