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

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## Chapter 7 - Study of Gas Laws Exercise Ex. 7

Question Num 1

What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 temperature remaining constant.

Solution Num 1

V1 = 500 dm3

P1 = 1 bar

T1 = 273 K

V2 = 500 dm3

T2 = 273 K

P2= ?

Question Num 2

2 litres of a gas is enclosed in a vessel at a pressure of 760 mmHg. If temperature remains constant, calculate pressure when volume changes to 4 dm3.

Solution Num 2

V = 2 litres

P = 760 mm

V1 = 4000 m3 [1 dm3 = 4 litres]

P1= ?

Question Num 3

At constant temperature, the effect of change of pressure on volume of a gas was as given below:

 Pressure in atmosphere Volume in litres 0.20 112 0.25 89.2 0.40 56.25 0.60 37.40 0.80 28.10 1.00 22.4

1. Plot the following graphs

1. P vs V

2. P vs 1/V

3. PV vs P

Interpret each graph in terms of a law.

1. Assuming that the pressure values given above are correct, find the correct measurement of the volume.
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

Question Num 4

800 cm3 of gas is collected at 650 mm pressure. At what pressure would the volume of the gas reduce by 40% of its original volume, temperature remaining constant?

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

Question Num 5

A cylinder of 20 litres capacity contains a gas at 100 atmospheric pressure. How many flasks of 200 cm3capacity can be filled from it at 1 atmosphere pressure, temperature remaining constant?

Solution Num 5

Question Num 6

A steel cylinder of internal volume 20 litres is filled with hydrogen at 29 atmospheric pressure. If hydrogen is used to fill a balloon at 1.25 atmospheric pressure at the same temperature, what volume will the gas occupy?

Solution Num 6

V = 20 litre

P = 29 atm

P1 = 1.25 atm

V1 =?

T = T1

Question Num 7

561 dm3 of a gas at STP is filled in a 748 dm3 container. If temperature is const

ant, calculate the percentage change in pressure required.

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 =

Question Num 8

88 cm3 of nitrogen is at a pressure of 770 mm mercury. If the pressure is raised to 880 mmHg, find by how much the volume will diminish, temperature remaining constant.

Solution Num 8

V = 88 cm3

P = 770 mm

P1 = 880 mm

V1= ?

T = T1

Volume diminishes = 88 - 77 = 11 cm3

Question Num 9

A gas at 240 K is heated to 127°C. Find the percentage change in the volume of the gas (pressure remaining constant).

Solution Num 9

Let volume = 100 ml

T = 240 K

Volume increased = x ml

New volume = 100 + x ml

T1 = 400 K

Question Num 10

Certain amount of a gas occupies a volume of 0.4 litre at 17°C. To what temperature should it be heated so that its volume gets (a) doubled, (b) reduced to half, pressure remaining constant?

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= ?

Question Num 11

A gas occupies 3 litres at 0°C. What volume will it occupy at -20°C, pressure remaining constant?

Solution Num 11

V = 3 litres

P = P1

V1= ?

T = 0°C = 0 + 273 = 273 K

T1 = -20°C = -20°C + 273 = 253 K

Question Num 12

A gas occupies 500 cm3 at normal temperature. At what temperature will the volume of the gas be reduced by 20% of its original volume, pressure being constant?

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

Question Num 13

Calculate the final volume of a gas 'X' if the original pressure of the gas at STP is doubled and its temperature is increased three times.

Solution Num 13

V1 = X

P1 = 1 atm

V2= ?

T2 = 3 T1

P2 = 2 atm

Question Num 14

A sample of carbon dioxide occupies 30 cm3 at 15°C and 740 mm pressure. Find its volume at STP.

Solution Num 14

V = 30 cm3

P = 740 mm

T = 288 K

P1 = 760 mm

V1= ?

T1 = 273 K

Question Num 15

50 cm3 of hydrogen is collected over water at 17°C and 750 mmHg pressure. Calculate the volume of a dry gas at STP. The water vapour pressure at 17°C is 14 mmHg.

Solution Num 15

V = 50 cm3

P = 750 - 14 = 736 mm

T = 290 K

P1 = 760 mm

V1= ?

T1 = 273 K

Question Num 16

At 0°C and 760 mmHg pressure, a gas occupies a volume of 100 cm3. Kelvin temperature of the gas is increased by one-fifth and the pressure is increased one and a half times. Calculate the final volume of the gas.

Solution Num 16

V = 100 cm3

P = 760 mm

T = 273 K

V1= ?

Question Num 17

It is found that on heating a gas its volume increases by 50% and its pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was heated.

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

Question Num 18

A certain mass of a gas occupies 2 litres at 27°C and 100 Pa. Find the temperature when volume and pressure become half of their initial values.

Solution Num 18

V = 2 litres

P = 100 Pa

T = 300 K

T1 = 75 - 273 = -198°C

Question Num 19

2500 cm3 of hydrogen is taken at STP. The pressure of this gas is further increased by two and a half times (temperature remaining constant). What volume will hydrogen occupy now?

Solution Num 19

V1 = 2500 cm3

P1 = 1 atm = 760 mm

T1 = 273 K

V2= ?

T2 = 273 K

Question Num 20

Taking the volume of hydrogen as calculated in Q.19, what change must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 (pressure remaining constant)?

Solution Num 20

V1 = 714.29 cm3

P1 = P2 = P

T1 = 273 K

V2 = 2500 cm3

T2= ?

Question Num 21

A given amount of gas A is confined in a chamber of constant volume. When the chamber is immersed in a bath of melting ice, the pressure of the gas is 100 cmHg.

a. What is the temperature when the pressure is 10 cmHg?

b. What will be the pressure when the chamber is brought to 100°C

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

Question Num 22

A gas is to be filled from a tank of capacity 10,000 litres into cylinders each having capacity of 10 litres. The condition of the gas in the tank is as follows:

1. Pressure inside the tank is 800 mmHg.
2. Temperature inside the tank is -3°C.

When the cylinder is filled, the pressure gauge reads 400 mmHg and the temperature is 0°C. Find the number of cylinders required to fill the gas.

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= ?

Question Num 23

Calculate the volume occupied by 2 g of hydrogen at 27°C and 4 atmosphere pressure if at STP it occupies 22.4 litres.

Solution Num 23

V1 = 22.4 litres

P1 = 1 atm

T1 = 273 K

V2 =?

T2 = 300 K

P2 = 4 atm

Question Num 24

What temperature would be necessary to double the volume of a gas initially at STP if the pressure is decreased to 50%?

Solution Num 24

V1 = V1

P1 = 760 atm

T1 = 273 K

V2 = 2V1

T2 =?

Question Num 25

Which will have greater volume when the following gases are compared at STP:

1. 1.2/N2 at 25°C and 748 mmHg
2. 1.25/O2 at STP
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.

Question Num 26

Calculate the volume of dry air at STP that occupies 28 cm3 at 14°C and 750 mmHg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mmHg.

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,

Question Num 27

An LPG cylinder can withstand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at 27°C. Because of a sudden fire in the building, the temperature rises. At what temperature will the cylinder explode?

Solution Num 27

P = 14.9 atm

V = 28 cm3

T = ?

P1 = 12 atm

V = V1

T1 = 300 K

Using gas equation,

Question Num 28

22.4 litres of a gas weighs 70 g at STP. Calculate the weight of the gas if it occupies a volume of 20 litres at 27°C and 700 mmHg of pressure.

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 =

Question 1

What do you understand by gas?

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.

Question 2

Give the assumptions of the kinetic molecular theory.

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.
Question 3

During the practical session in the lab when hydrogen sulphide gas having offensive odour is prepared for some test, we can smell the gas even 50 metres away. Explain the phenomenon.

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.

Question 4

What is diffusion? Give an example to illustrate it.

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.

Question 5

How is molecular motion related with temperature?

Solution 5

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

Question 6

State (i) the three variables for gas laws and (ii) SI units of these variables.

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)

Question 7
1. State Boyle's Law.
2. Give its

i. Mathematical expression

ii. Graphical representation and

iii. Significance

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.

Question 8

Explain Boyle's Law on the basis of the kinetic theory of matter.

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.

Question 9

The molecular theory states that the pressure exerted by a gas in a closed vessel results from the gas molecules striking against the walls of the vessel. How will the pressure change if

1. The temperature is doubled keeping the volume constant
2. The volume is made half of its original value keeping the T constant
Solution 9
1. Pressure will double.
2. Pressure remains the same.
Question 10
1. State Charles's law.
2. Give its

i. Graphical representation

ii. Mathematical expression and

iii. Significance

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.

Question 11

Explain Charles's law on the basis of the kinetic theory of matter.

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).

Question 12

Define absolute zero and absolute scale of temperature. Write the relationship between °C and K.

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

Question 13
1. What is the need for the Kelvin scale of temperature?
2. What is the boiling point of water on the Kelvin scale? Convert it into centigrade scale.
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.

Question 14
1. Define STP or NTP.
2. Why is it necessary to compare gases at STP?
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.
Question 15

Correct the following statements:

1. Volume of a gas is inversely proportional to its pressure at constant temperature.
2. Volume of a fixed mass of a gas is directly proportional to its temperature, pressure remaining constant.
3. 0°C is equal to zero Kelvin.
4. Standard temperature is 25°C.
5. Boiling point of water is 273 K.
Solution 15
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.
Question 16
1. What is the relationship between the Celsius and Kelvin scales of temperature?
2. Convert (i) 273°C to Kelvin and (ii) 293 K to °C.
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

Question 17

State the laws which are represented by the following graphs:

Solution 17
1. Charles's law
2. Boyle's law
Question 18

Give reasons for the following:

1. All temperature in the absolute (Kelvin) scale are in positive figures.
2. Gases have lower density compared to solids or liquids.
3. Gases exert pressure in all directions.
4. It is necessary to specify the pressure and temperature of a gas while stating its volume.
5. Inflating a balloon seems to violate Boyle's law.
6. Mountaineers carry oxygen cylinders with them.
7. Gas fills the vessel completely in which it is kept.
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.
Question 19

How did Charles's law lead to the concept of absolute scale of temperature?

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.

Question 20

What is meant by aqueous tension? How is the pressure exerted by a gas corrected to account for aqueous tension?

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

Question 21

State the following:

1. Volume of a gas at 0 Kelvin
2. Absolute temperature of a gas at 7°C
3. Gas equation
4. Ice point in absolute temperature
5. STP conditions
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.

Question 22

1. The graph of PV vs P for a gas is
1. Parabolic
2. Hyperbolic
3. A straight line parallel to the X-axis
4. A straight line passing through the origin

1. The absolute temperature value that corresponds to 27°C is
1. 200 K
2. 300 K
3. 400 K
4. 246 K

1. Volume-temperature relationship is given by
1. Boyle
2. Gay-Lussac
3. Dalton
4. Charles

1. If pressure is doubled for a fixed mass of a gas, its volume will become
1. 4 times
2. ½ times
3. 2 times
4. No change
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
Question 23

Match the following:

 Column A Column B (a) cm3 (i) Pressure (b) Kelvin (ii) Temperature (c) Torr (iii) Volume (d) Boyle's law (a) Charles's law

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

Question 24

Write the value of

1. Standard temperature in
1. °C
2. K
1. Standard pressure in
1. atm
2. mmHg
3. cmHg
4. torr
Solution 24
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
Question 25
1. The average kinetic energy of the molecules of a gas is proportional to the ………….
2. The temperature on the Kelvin scale at which molecular motion completely ceases is called……………
3. If temperature is reduced to half, ………….. would also reduce to half.
4. The melting point of ice is …………. Kelvin.
Solution 25
1. Absolute temperature
2. Absolute zero
3. Volume
4. 273