SELINA Solutions for Class 9 Chemistry Chapter 7  Study of Gas Laws
Chapter 7  Study of Gas Laws Exercise Ex. 7
What will be the minimum pressure required to compress 500 dm^{3} of air at 1 bar to 200 dm^{3} temperature remaining constant.
V_{1} = 500 dm^{3}
P_{1} = 1 bar
T_{1} = 273 K
V_{2} = 500 dm^{3}
T_{2} = 273 K
P_{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 dm^{3}.
V = 2 litres
P = 760 mm
V_{1} = 4000 m^{3} [1 dm^{3} = 4 litres]
P_{1}= ?
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 
 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.
 Assuming that the pressure values given above are correct, find the correct measurement of the volume.
P/atm 
V/dm^{3} 
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.
 The correct measurements of the volume are given below:
P/atm 
V/dm^{3} 
0.2 
112 
0.25 
89.6 
0.4 
56 
0.6 
37.33 
0.8 
28 
1 
22.4 
800 cm^{3} 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?
Given:
V = 800 cm^{3}
P = 650 m
P_{1}= ?
V_{1} = reduced volume = 40% of 800
=
Net V_{1} = 800  320 = 480 cm^{3}
T = T_{1}
Using the gas equation,
Since T = T_{1}
A cylinder of 20 litres capacity contains a gas at 100 atmospheric pressure. How many flasks of 200 cm^{3}capacity can be filled from it at 1 atmosphere pressure, temperature remaining constant?
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?
V = 20 litre
P = 29 atm
P_{1} = 1.25 atm
V_{1} =?
T = T_{1}
561 dm^{3} of a gas at STP is filled in a 748 dm^{3} container. If temperature is const
ant, calculate the percentage change in pressure required.
Initial volume = V_{1} = 561 dm^{3}
Final volume = V_{2} = 748 dm^{3}
Difference in volume = 748  561 = 187 dm^{3}
As the temperature is constant,
Decrease in pressure percentage =
88 cm^{3} 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.
V = 88 cm^{3}
P = 770 mm
P_{1} = 880 mm
V_{1}= ?
T = T_{1}
Volume diminishes = 88  77 = 11 cm^{3}
A gas at 240 K is heated to 127°C. Find the percentage change in the volume of the gas (pressure remaining constant).
Let volume = 100 ml
T = 240 K
Volume increased = x ml
New volume = 100 + x ml
T_{1} = 400 K
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?
(a) V_{1} = 0.4 L
V_{2} = 0.4 × 2L
T_{1} = 17°C (17 + 273) = 290 K
T_{2}= ?
(b) V_{1} = 0.4 L
V_{2} = 0.2 L
T_{1} = 17°C (17 + 273) = 290 K
T_{2}= ?
A given mass of a gas occupies 143 cm^{3} at 27^{o}C and 700mm Hg pressure. What will be its volume at 300K and 280 mm Hg pressure?
V_{1} = 143 cm^{3}
T_{1} = 27^{o}C = 27+273 = 300K
P_{1} = 700 mm Hg
T_{2} = 300K
P_{2}= 280 mm Hg
V_{2}= ?
_{2} = 357.5 cm^{3}
A gas occupies 500 cm^{3} at normal temperature. At what temperature will the volume of the gas be reduced by 20% of its original volume, pressure being constant?
V = 500 cm^{3 }
Normal temperature, t = 0°C = 0 + 273 K
V_{1} = Reduced volume + 20% of 500 cm^{3}
Net, V_{1 } = 500  100 = 400 cm^{3}
T_{1}= ?
P = P_{1}
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.
V_{1} = X
P_{1} = 1 atm
V_{2}= ?
T_{2} = 3 T_{1}
P_{2} = 2 atm
A sample of carbon dioxide occupies 30 cm^{3} at 15°C and 740 mm pressure. Find its volume at STP.
V = 30 cm^{3}
P = 740 mm
T = 288 K
P_{1} = 760 mm
V_{1}= ?
T_{1} = 273 K
What temperature would be necessary to double the volume of a gas initially at STP if the pressure is decreased to 50%?
V_{1} = V_{1}
P_{1} = 760 atm
T_{1} = 273 K
V_{2} = 2V_{1}
T_{2} =?
At 0°C and 760 mmHg pressure, a gas occupies a volume of 100 cm^{3}. Kelvin temperature of the gas is increased by onefifth and the pressure is increased one and a half times. Calculate the final volume of the gas.
V = 100 cm^{3}
P = 760 mm
T = 273 K
V_{1}= ?
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.
Let the original volume (V) = 1 and
the original pressure (P) = 1 and
the temperature given (T) = 15°C = 15 + 273 = 258 K
V_{1 }or new volume after heating = original volume + 50% of original volume
P_{1} or decreased pressure = 60%
T_{1} = to be calculated
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.
V = 2 litres
P = 100 Pa
T = 300 K
T_{1} = 75  273 = 198°C
2500 cm^{3} 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?
V_{1} = 2500 cm^{3}
P_{1} = 1 atm = 760 mm
T_{1} = 273 K
V_{2}= ?
T_{2} = 273 K
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 cm^{3} (pressure remaining constant)?
V_{1} = 714.29 cm^{3}
P_{1} = P_{2} = P
T_{1} = 273 K
V_{2} = 2500 cm^{3}
T_{2}= ?
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
 V_{1} = V_{2} = V
P_{1} = 100 cmHg
T_{1} = 273 K
P_{2} = 10 cmHg
T_{2}= ?
 V_{1} = V_{2} = V
P_{1} = 100 cmHg
P_{2}= ?
T_{1} = 273 K
T_{2} = 373 K
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:
 Pressure inside the tank is 800 mmHg.
 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.
Capacity of the cylinder V = 10000 litres
P = 800 mm
T = 3°C = 3 + 273 = 270 K P_{1} = 400 mmHg
T_{1} = 0°C = 0 + 273 = 273 K
V_{1}= ?
Calculate the volume occupied by 2 g of hydrogen at 27°C and 4 atmosphere pressure if at STP it occupies 22.4 litres.
V_{1} = 22.4 litres
P_{1} = 1 atm
T_{1} = 273 K
V_{2} =?
T_{2} = 300 K
P_{2} = 4 atm
50 cm^{3} 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.
V = 50 cm^{3}
P = 750  14 = 736 mm
T = 290 K
P_{1} = 760 mm
V_{1}= ?
T_{1} = 273 K
Which will have greater volume when the following gases are compared at STP:
 1.2/N_{2} at 25°C and 748 mmHg
 1.25/O_{2} at STP
 V = 1.2 litres
P = 748 mmHg
T = 298 K
P_{1} = 760 mmHg
T_{1} = 273 K
V_{1}= ?
 V = 1.25 litres
P = 760 mmHg
T = 273 K
P_{1} = 760 mmHg
T_{1} = 273 K
V_{1}= ?
1.25 litres O_{2} will have greater volume than 1.2 litres N_{2}.
Calculate the volume of dry air at STP that occupies 28 cm^{3 }at 14°C and 750 mmHg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mmHg.
Pressure due to dry air,
P = 750  12 = 738 mm
V = 28 cm^{3}
T = 14°C = 14 + 273 = 287 K
P_{1} = 760 mmHg
V_{1}= ?
T_{1} = 0°C = 273 K
Using gas equation,
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?
P = 14.9 atm
V = 28 cm^{3}
T = ?
P_{1} = 12 atm
V = V_{1}
T_{1} = 300 K
Using gas equation,
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.
Step 1:
V_{1} = 20 litres
P_{1} = 700 mm
T_{1} = 300 K
V_{2}= ?
T_{2} = 273 K
P_{2} = 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 =
What do you understand by gas?
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.
Give the assumptions of the kinetic molecular theory.
 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.
 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.
 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.
 There is large interparticle space between gas molecules, and this accounts for high compressibility of gases.
 Volume of a gas increases with a decrease in pressure and increase in temperature.
 Gases have low density as they have large intermolecular spaces between their molecules.
 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.
 The intermolecular space of a gas is reduced because of cooling. Molecules come closer resulting in liquefaction of the gas.
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.
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.
What is diffusion? Give an example to illustrate it.
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.
How is molecular motion related with temperature?
Temperature affects the kinetic energy of molecules. So, molecular motion is directly proportional to temperature.
State (i) the three variables for gas laws and (ii) SI units of these variables.
 Three variables for gas laws: Volume (V), Pressure (P), Temperature (T)
 SI units of these variables:
For volume: Cubic metre (m^{3})
For pressure: Pascal (Pa)
For temperature: Kelvin (K)
 State Boyle's Law.
 Give its
i. Mathematical expression
ii. Graphical representation and
iii. Significance
 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 Xaxis 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.
Explain Boyle's Law on the basis of the kinetic theory of matter.
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 onehalf 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.
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
 The temperature is doubled keeping the volume constant
 The volume is made half of its original value keeping the T constant
 Pressure will double.
 Pressure remains the same.
 State Charles's law.
 Give its
i. Graphical representation
ii. Mathematical expression and
iii. Significance
Charles's Law
At constant pressure, the volume of a given mass of a dry gas increases or decreases by 1/273^{rd} of its original volume at 0°C for each degree centigrade rise or fall in temperature.
V ∝ T (at constant pressure) 
At temperature T_{1} (K) and volume V_{1} (cm^{3}):
...(i)
At temperature T_{2} (K) and volume V_{2} (cm^{3}):
….(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.
Explain Charles's law on the basis of the kinetic theory of matter.
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).
Define absolute zero and absolute scale of temperature. Write the relationship between °C and K.
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
 What is the need for the Kelvin scale of temperature?
 What is the boiling point of water on the Kelvin scale? Convert it into centigrade scale.
 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.
 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.
 Define STP or NTP.
 Why is it necessary to compare gases at STP?
 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) 
 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.
Write the value of
 Standard temperature in
 °C
 K
 Standard pressure in
 atm
 mmHg
 cmHg
 torr
 °C = O°C
 K = 273 K
 1 atm
 760 mmHg
 76 cmHg
 1 torr = 133.32 Pascal
 What is the relationship between the Celsius and Kelvin scales of temperature?
 Convert (i) 273°C to Kelvin and (ii) 293 K to °C.
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
State the laws which are represented by the following graphs:
 Charles's law
 Boyle's law
Give reasons for the following:
 All temperature in the absolute (Kelvin) scale are in positive figures.
 Gases have lower density compared to solids or liquids.
 Gases exert pressure in all directions.
 It is necessary to specify the pressure and temperature of a gas while stating its volume.
 Inflating a balloon seems to violate Boyle's law.
 Mountaineers carry oxygen cylinders with them.
 Gas fills the vessel completely in which it is kept.
 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.
 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.
 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.
 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.
 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.
 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.
 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.
How did Charles's law lead to the concept of absolute scale of temperature?
 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.
 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.
 The temperature 273°C is called absolute zero.
What is meant by aqueous tension? How is the pressure exerted by a gas corrected to account for aqueous tension?
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 
P_{total} = P_{gas} + P_{water}_{ vapour }
P_{gas} = P_{total} P_{water}_{ vapour}
Actual pressure of gas = Total pressure  Aqueous tension
State the following:
 Volume of a gas at 0 Kelvin
 Absolute temperature of a gas at 7°C
 Gas equation
 Ice point in absolute temperature
 STP conditions
 Volume of gas is zero.
 Absolute temperature is 7 + 273 = 280 K.
 The gas equation is
 Ice point = 0 + 273 = 273 K
 Standard temperature is taken as 273 K or O°C.
Standard pressure is taken as 1 atmosphere (atm) or 760 mmHg.
Choose the correct answer:
 The graph of PV vs P for a gas is
 Parabolic
 Hyperbolic
 A straight line parallel to the Xaxis
 A straight line passing through the origin
 The absolute temperature value that corresponds to 27°C is
 200 K
 300 K
 400 K
 246 K
 Volumetemperature relationship is given by
 Boyle
 GayLussac
 Dalton
 Charles
 If pressure is doubled for a fixed mass of a gas, its volume will become
 4 times
 ½ times
 2 times
 No change
 (iii) Straight line parallel to the Xaxis.
 (ii) 27°C = 27 + 273 = 300 K
 (iv) Charles
 (ii) 1/2 times
Match the following:

Column A 
Column B 
(a) 
cm^{3} 
(i) Pressure 
(b) 
Kelvin 
(ii) Temperature 
(c) 
Torr 
(iii) Volume 
(d) 
Boyle's law 

(a) 
Charles's law 




Column A 
Column B 
(a) cm^{3} 
Volume 
(b) Kelvin 
Temperature 
(c) Torr 
Pressure 
(d) Boyle's law 
PV = P_{1} V_{1}

(e) Charles's law 

Correct the following statements:
 Volume of a gas is inversely proportional to its pressure at constant temperature.
 Volume of a fixed mass of a gas is directly proportional to its temperature, pressure remaining constant.
 0°C is equal to zero Kelvin.
 Standard temperature is 25°C.
 Boiling point of water is 273 K.
 Volume of a gas is directly proportional to the pressure at constant temperature.
 Volume of a fixed mass of a gas is inversely proportional to the temperature, the pressure remaining constant.
 273°C is equal to zero Kelvin.
 Standard temperature is 0°C.
 The boiling point of water is 373 K.
 The average kinetic energy of the molecules of a gas is proportional to the ………….
 The temperature on the Kelvin scale at which molecular motion completely ceases is called……………
 If temperature is reduced to half, ………….. would also reduce to half.
 The melting point of ice is …………. Kelvin.
 Absolute temperature
 Absolute zero
 Volume
 273
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