Chapter7 Study of Gas Laws Class9 Concise Selina Solutions . Step by Step Selina Concise Solutions for ICSE Class9 Chemistry of Chapter7 Exercise Questions . The Solutions of Numerical Based on Boyel’s Law, (Page 119) , Numerical Based on Charles’ Law, (Page 122) , Exercise7 for Selina Concise Chemistry ICSE Class9. Visit official website cisce for detail information about icse class 9 Chemistry.
Chapter7 Study of Gas Laws Class9 Concise Selina Solutions
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Numerical Based on Boyel’s Law, (Page 119)
Numerical Based on Charles’ Law, (Page 122)
Numerical Based on Boyel’s Law, (Page 119) Study of Gas Laws Chapter7 Selina Chemistry Solutions Class 9
Question 1
Volume of certain amount of a gas at 25°C and 100 cm Hg pressure is 80 mL. The gas is expanded to 160 mL keeping the temperature constant. Calculate the pressure of the expanded gas.
Answer 1
Question 2
At a particular temperature, a certain quantity of gas occupies a volume of 74 cm at a pressure of 760 mm. If the pressure is decreased to 740 mm, what will be the volume of the gas at the same temperature ?
Answer 2
Questions 3
L A student performed an experiment to measure pressure and volume of a gas at constant temperature and noted the following :
……………
Calculate the value of x and y. Which law was used in the calculations ? Draw a suitable graph.
Answer 3
Question 4.
At a constant temperature, volume of a gas was found to be 400 cm’ at a pressure of 760mm Hg. If the pressure of the gas is increased by 25%, find the new volume.
Answer 4
Question 5.
A vessel of capacity 600 cm³ contains hydrogen gas at a pressure of 330 cm Hg. What will be the pressure of hydrogen gas, when the vessel is connected to another vessel of 300 cm capacity ?
Answer 5
Questions 6
At constant temperature, a gas is at a pressure of 1080 mm Hg. If the volume is decreased by 40%, find the new pressure of the gas.
Answer 6
Numerical Based on Charles’ Law, (Page 122) Study of Gas Laws Chapter7 Selina Chemistry Solutions Class 9
Question 1.
Convert the following :
(i) 37 K to °C
(ii) 273 K to °C
(iii) 27°C toK
(iv) 27°C to K
Answer 1
Question 2.
20 mL of hydrogen gas at 13°C is heated to 117°C at constant pressure. Find the new volume of hydrogen.
Answer 2
Question 3.
At what temperature in degree centigrade will the volume of a gas, which is originally at 0°C, double itself, pressure remaining constant.
Answer 3
Question 4.
Calculate the volume (in cm³) of air expelled from a vessel containing 04 litres of it at 250 K, when it is heated to 27°C at the same pressure.
Answer 4
Question 5.
What will be the volume of a gas when 3 litres of it is cooled down from 27°C to 73°C at constant pressure.
Answer 5
Question 6.
To what temperature must a gas at 300 K be cooled down in order to reduce its volume to 1/3rd of its original volume, pressure remaining
constant ?
Answer 6
Question 7
Prove that the volume of a gas at 273°C is twice its volume at 273 K, at constant pressure.
Answer 7
Question 8.
A gas occupies 3 litres at 0°C. What volume will it occupy at 20°C, pressure remaining
Answer 8
Exe7 Study of Gas Laws Chapter7 Selina Chemistry Solutions Class 9
Question 1
What do you understand by gas?
Answer 1
Gas is a state of matter in which inter particle attraction is weak and inter particle 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.
Answer 2
 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 inter particle space between gas molecules, and this accounts for high comprehensibility of gases.
 Volume of a gas increases with a decrease in pressure and increase in temperature.
 Gases have low density as they have large inter molecular spaces between their molecules.
 Gases have a natural tendency to mix with one and other because of large inter molecular spaces. So, two gases when mixed form a homogeneous gaseous mixture.
 The inter molecular space of a gas is reduced because of cooling. Molecules come closer resulting in liquefaction of the gas.
Question 3
During the practical 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.
Answer 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.
Answer 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?
Answer 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.
Answer 6
(i) Three variables for gas laws: Volume (V), Pressure (P), Temperature (T)
(ii) SI units of these variables:
For volume: Cubic meter (m^{3})
and For pressure: Pascal (Pa)
For temperature: Kelvin (K)
Question 7
(a) State Boyle’s Law.
(b) Give its
(i) Mathematical expression
(ii) Graphical representation and
(iii). Significance
Answer 7
(a) 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
(a) Mathematical representation:
(i)
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.
Answer 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 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.
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
(a) The temperature is doubled keeping the volume constant
(b) The volume is made half of its original value keeping the T constant
Answer 9
(a) Pressure will double.
(b) Pressure remains the same.
Question 10
(a) State Charles’s law.
(b) Give its
(i)Graphical representation
(ii) Mathematical expression and
(iii) . Significance
Answer 10
(a) 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
(b) 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.
Answer 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.
Answer 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
(a) What is the need for the Kelvin scale of temperature?
(b) What is the boiling point of water on the Kelvin scale? Convert it into centigrade scale.
Answer 13
(a) The behavior 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.
(b) 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
(a) Define STP.
(b) Why is it necessary to compare gases at STP?
Answer 14
(a) 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 Temperature 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) 
(b) 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 16
(a) what is the relationship between the Celsius and the Kelvin scales of temperature ?
(b) Convert (i) 273°C to Kelvin (ii) 293 K to °C
Answer 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:
Answer 17
 Charles’s law
 Boyle’s law
Question 18
Give reasons for the following:
(a) All temperature in the absolute (Kelvin) scale are in positive figures.
(b) Gases have lower density compared to solids or liquids.
(c) Gases exert pressure in all directions.
(d) It is necessary to specify the pressure and temperature of a gas while stating its volume.
(e) Inflating a balloon seems to violate Boyle’s law.
(f) Mountaineers carry oxygen cylinders with them.
(g) Gas fills the vessel completely in which it is kept.
Answer 18
(a) 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.
(b) 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.
(c) 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.
(d) 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.
(e) 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.
(f) 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.
(g) 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?
Answer 19
 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.
Question 20
What is meant by aqueous tension? How is the pressure exerted by a gas corrected to account for aqueous tension?
Answer 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 
P_{total} = P_{gas} + P_{water}_{ vapour}
P_{gas} = P_{total}– P_{water}_{ vapour}
Actual pressure of gas = Total pressure – Aqueous tension
Question 21
State the following:
(a) Volume of a gas at 0 Kelvin
(b) Absolute temperature of a gas at 7°C
(c) Gas equation
(d) Ice point in absolute temperature
(e) STP conditions
Answer 21
(a) Volume of gas is zero.
(b) Absolute temperature is 7 + 273 = 280 K.
(c) The gas equation is
(d) Ice point = 0 + 273 = 273 K
(e) Standard temperature is taken as 273 K or O°C.
Standard pressure is taken as 1 atmosphere (atm) or 760 mmHg.
Question 22
Choose the correct answer:
(a) The graph of PV vs P for a gas is
(i) Parabolic
(ii) Hyperbolic
(iii) A straight line parallel to the Xaxis
(iv) A straight line passing through the origin
(b) The absolute temperature value that corresponds to 27°C is
(i) 200 K
(ii) 300 K
(iii) 400 K
(iv) 246 K
(c) Volumetemperature relationship is given by
(i) Boyle
(ii) GayLussac
(iii) Dalton
(iv) Charles
(d) If pressure is doubled for a fixed mass of a gas, its volume will become
(i) 4 times
(ii) ½ times
(iii) 2 times
(iv) No change
Answer 22
(a) (iii) Straight line parallel to the Xaxis.
(b) (ii) 27°C = 27 + 273 = 300 K
(c) (iv) Charles
(d) (ii) 1/2 times
Question 23
Match the following:
Column A  Column B  
(a)  cm^{3}  (i) Pressure 
(b)  Kelvin  (ii) Temperature 
(c)  Torr  (iii) Volume 
(d)  Boyle’s law  
(e)  Charles’s law  
Answer 23
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 
Question 24
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.
Answer 24
 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.
Question 25
Fill in the blanks
 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.
Answer 25
 Absolute temperature
 Absolute zero
 Volume
 273
Numerical Problems Chapter – 7 Study of Gas Law Selina Chemistry Solutions
Question 1
What will be the minimum pressure required to compress 500 dm^{3} of air at 1 bar to 200 dm^{3} temperature remaining constant.
Answer 1
V_{1} = 500 dm^{3}
P_{1} = 1 bar
T_{1} = 273 K
V_{2} = 500 dm^{3}
T_{2} = 273 K
P_{2}= ?
Question 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}.
Answer 2
V = 2 litres
P = 760 mm
V_{1} = 4000 m^{3} [1 dm^{3} = 4 litres]
P_{1}= ?
Question 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 
(a) Plot the following graphs
 P vs V
 P vs 1/V
 PV vs P
Interpret each graph in terms of a law.
(b) Assuming that the pressure values given above are correct, find the correct measurement of the volume.
Answer 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 
(a) Plot the following graphs
 P vs V
2. P vs 1/V
3. PV vs P
Interpret each graph in terms of a law.
(b) Assuming that the pressure values given above are correct, find the correct measurement of the volume.
Question 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?
Answer 4
Given:
V = 800 cm^{3}
P = 650 m
P_{1}= ?
V_{1} = reduced volume = 40% of 800
= 8000× 40/100= 320
Net V_{1} = 800 – 320 = 480 cm^{3}
T = T_{1}
Using the gas equation,
Since T = T_{1}
=1083.33 mm of Hg.
Question 5 Study of Gas Laws Chapter7
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?
Answer 5
Question 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?
Answer 6
V = 20 litre
P = 29 atm
P_{1} = 1.25 atm
V_{1} =?
T = T_{1}
Question 7
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.
Answer 7
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 =
(187×100)/748 = 25%
Question 8
88 cm^{3} of nitrogen is at a pressure of 770 mm mercury. If the pressure is raised to 880 mm Hg, find by how much the volume will diminish, temperature remaining constant.
Answer 8
V = 88 cm^{3}
P = 770 mm
P_{1} = 880 mm
V_{1}= ?
T = T_{1}
Using gas equation,
Volume diminishes = 88 – 77 = 11 cm^{3}
Question 9
A gas at 240 K is heated to 127°C. Find the percentage change in the volume of the gas (pressure remaining constant).
Answer 9
Let volume = 100 ml
T = 240 K
Volume increased = x ml
New volume = 100 + x ml
T_{1} = 400 K
Question 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?
Answer 10
(a) V_{1} = 0.4 L
V_{2} = 0.4 × 2L
T_{1} = 17°C (17 + 273) = 290 K
T_{2}= ?
=145 – 273 = 128°C
Question 11
A given mass of a gas occupies 143 cm³ at 27°C and 700 mm of Hg pressure. What will be its volume at 300 K and 280 mm of Hg pressure?
Answer 11
Question 12
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?
Answer 12
V = 500 cm^{3 }
Normal temperature, t = 0°C = 0 + 273 K
V_{1} = Reduced volume + 20% of 500 cm^{3}
Question 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.
Answer 13
Question 14
A sample of carbon dioxide occupies 30 cm^{3} at 15°C and 740 mm pressure. Find its volume at STP.
Answer 14
Question 15
What temperature would be necessary to double the volume of a gas initially at STP if the pressure is decreased to 50%?
Answer 15
V_{1} = V_{1}
P_{1} = 760 atm
T_{1} = 273 K
V_{2} = 2V_{1}
T_{2} =?
Question 16
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.
Answer 16
Question 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.
Answer 17
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
Question 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.
Answer 18
Question 19
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?
Answer 19
Question 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 cm^{3} (pressure remaining constant)?
Answer 20
Question 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 cm Hg.
(a) What is the temperature when the pressure is 10 cmHg?
(b) What will be the pressure when the chamberis brought to 100°C
Answer 21
Question 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:
(a) Pressure inside the tank is 800 mmHg.
(b) 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.
Answer 22
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}= ?
Question 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.
Answer 23
V_{1} = 22.4 litres
P_{1} = 1 atm
T_{1} = 273 K
V_{2} =?
T_{2} = 300 K
P_{2} = 4 atm
Question 24
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 mm Hg.
Answer 24
Question 25
Which will have greater volume when the following gases are compared at STP:
(a) 1.2/N_{2} at 25°C and 748 mmHg
(b) 1.25/O_{2} at STP
Answer 25
Question 26
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.
Answer 26
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,
Question 27
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?
Answer 27
P = 14.9 atm
V = 28 cm^{3}
T = ?
P_{1} = 12 atm
V = V_{1}
T_{1} = 300 K
Using gas equation,
Question 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.
Amswer 28
Step 1:
V_{1} = 20 litres
P_{1} = 700 mm
T_{1} = 300 K
V_{2}= ?
T_{2} = 273 K
P_{2} = 760 mm
–: End of Study of Gas Laws Chapter7 Selina Chemistry Solutions Class 9 :–
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