Selina Concise Solutions for ICSE Class 9 Chemistry Chapter 2 Study Of Gas Laws

Exercise 2

Question 1:
Answer: The 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, is known as gas.
In a gas, the atoms or molecules have so much energy that they overcome any attraction to stay together. This explains why gases do not have a fixed shape and will expand to fill any container you put them in.
Teacher's Tip: Think of gas particles like energetic toddlers running around a huge playground.
Exam Tip: Always mention "random motion" and "entire available space" when defining the gaseous state for full marks.

Question 2:
Answer: The main assumption of kinetic molecular theory of gases are as follows:
1. All gases are made up of a large number of extremely small particles called molecules.
2. There are large vacant spaces between the molecules of a gas so that actual volume of the molecules of a gas is negligible as compared to the total volume occupied by the gas.
3. The molecules of a gas are always in a state of constant random motion in straight lines in all possible direction.
4. There are negligible attractive forces between the molecules of a gas.
5. There is no effect of gravity on the motion of the molecules of a gas.
6. The average kinetic energy of the molecules of a gas is directly proportional to that of the Kelvin temperature of the gas.
7. The molecules are perfectly elastic so that there is no net loss of energy during molecular collisions.
8. The pressure of a gas is due to the bombardment of the molecules of a gas against the walls of a container.
This theory provides a mental model of how invisible gas particles behave like tiny, bouncy balls. By assuming the particles are "perfectly elastic," scientists can calculate gas behavior without worrying about energy being lost as heat during every tiny collision.
Teacher's Tip: Remember point 8 - pressure is just the result of trillions of tiny "punches" against the container walls.
Exam Tip: If asked for the relationship between temperature and energy, remember it is always "directly proportional" to the Kelvin scale.

Question 3:
Answer: In a laboratory, when hydrogen sulphide gas is prepared, it can be smelt even at 50 meters away. This is due to the phenomenon called Diffusion.
Diffusion is a process of intermixing of two substances kept in contact.
The inter-particle or inter molecular spaces in a gas are very large. When hydrogen sulphide gas is produced, its particle collides with air particles. Due to the collisions of particles, they start moving in all possible directions. As a result of which the two gases mix with each other forming a homogeneous mixture of a gas. Thus, the released gas can be smelt to a long distance.
Diffusion happens fastest in gases because the particles are moving very quickly and there is plenty of empty space to move through. This is why smells like perfume or cooking food spread through a house even without a breeze.
Teacher's Tip: Diffusion is like "automatic mixing" that happens all by itself.
Exam Tip: To explain diffusion, always mention "large intermolecular spaces" and "random motion" of particles.

Question 4:
Answer: Pressure and volume relationship of gases -
Experiment: Take a 10 ml syringe fitted with a piston. Raise the latter to the 10 ml mark and wrap an adhesive tape over its nozzle. Fit the wrapped nozzle tightly into a hole, bored half way through a rubber stopper.
Observation: On placing some weight on the piston (to put pressure), the piston moves downward and reduces the volume of air. Gradually, put more weight. The piston moves further downward and the volume of the air is further reduced.
Now remove the weights one by one. You will notice that, on decreasing the pressure, the piston moves upward as such the volume of the air increases.
Conclusion:
1. An increase in pressure at constant temperature causes a decrease in the volume of a gas; conversely, if the volume of a fixed mass of a gas at constant temperature is decreased, the pressure of the gas increases.
2. A decrease in pressure at constant temperature causes a increase in the volume of a gas; conversely, if the volume of a fixed mass of a gas at constant temperature is increased, the pressure of the gas decreases.
This experiment demonstrates that gases are highly compressible, which is a unique physical property compared to solids or liquids. When you push the plunger, you aren't shrinking the molecules; you are simply squeezing the empty space between them.
Teacher's Tip: Imagine squeezing a sponge; you are pushing the air out of the holes to make the whole thing smaller.
Exam Tip: Be sure to state that "temperature must remain constant" for this relationship to be true.

Question 5:
Answer: The molecular motion is directly proportional to the temperature.
As temperature increases, molecular motion increases because the molecule possesses certain kinetic energy. And as the temperature decreases, molecular motion also decreases. Thus, when temperature is zero, molecular motion stops or ceases.
Temperature is actually just a measurement of how fast molecules are vibrating or flying around. When you heat a gas, you are giving the particles more "fuel" to move faster and hit things harder.
Teacher's Tip: Heat is the "speed limit" for molecules; the hotter it is, the faster they are allowed to go!
Exam Tip: Use the term "Kinetic Energy" to describe the energy of motion for better marks.

Question 6:
Answer: The three variables for gas laws are:
1. Volume, V
2. Pressure, P
3. Temperature, T
These three are called as the Standard variables. S.I. unit of volume is cubic meter (m³). S.I. unit of pressure is Pascal (Pa). S.I. unit of temperature is Kelvin (K) or degree Celsius (°C).
These three variables are like the "Big Three" of gas physics; if you change one, at least one of the others will react. Scientists use these standard units to make sure everyone in the world is talking about the same measurements.
Teacher's Tip: Remember "V-P-T" like the name of a VIP club for gases.
Exam Tip: Always convert Celsius to Kelvin by adding 273 before using temperature in gas law formulas.

Question 7:
Answer: 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.
Mathematical representation:
According to Boyle’s Law,
V ∝ 1/P
V = K . 1/P
= K/P
K = VP or PV
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,
V' ∝ {1}/{P'}
V' = K . 1/P
= K/P
K = V'P' or PV
Graphical representation of Boyle's Law:
1. V vs 1/P: 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 X-axis is obtained.
Boyle's Law describes the "Goldilocks" relationship between pressure and volume: as one goes up, the other must go down to keep the balance. The constant value 'K' proves that the "strength" of the gas remains the same as long as it doesn't get hotter or colder.
Teacher's Tip: "Inversely proportional" just means they are on a seesaw; when Pressure goes up, Volume goes down!
Exam Tip: When drawing the V vs P graph, make sure the line is a curve (hyperbola) that never actually touches the axes.

Significance of Boyles law:
Answer: According to Boyles 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 that the mountaineers have to carry oxygen cylinders with them.
This law explains why air feels "thin" on top of a mountain; there isn't enough pressure to squeeze the oxygen molecules together into a dense, breathable space. Understanding this helps engineers design airplanes and life-saving equipment for high-altitude travel.
Teacher's Tip: Think of it as "squashed air" at sea level vs "loose air" on a mountain.
Exam Tip: Link "low pressure" to "low density" to explain why breathing is harder at high altitudes.

Question 8:
Answer: Explanation of Boyle’s Law on the basis of kinetic theory of matter.
According to kinetic theory of matter, the number of particles present in a given mass and the average kinetic energy is constant.
If the volume of given mass of a gas is reduced to half of its original volume. The same number of particles will have half space to move.
As a result, the number of molecules striking the unit area of the walls of the container at given time will get doubled of the pressure will also get doubled.
Alternatively, if the volume of a given mass of a gas is doubled at constant temperature, same number of molecules will have double space to move. Thus, number of molecule striking the unit area of the walls of container at a given time will become one half of original value. Thus, pressure will also get reduced to half of original pressure. Hence, it is seen that if pressure increases, volume of a gas decreases at constant temperature and this is Boyle’s law.
Pressure is simply the total force of molecules hitting the walls; if you shrink the room, those molecules hit the walls more often. This microscopic view proves that the law isn't just a random rule, but a logical result of how particles move.
Teacher's Tip: It's like putting the same 20 people in a small elevator vs. a large ballroom; they'll bump into the walls much more in the elevator!
Exam Tip: Use the phrase "number of collisions per unit area" to describe how pressure increases when volume decreases.

Question 9:
Answer: (a) Pressure will be doubled. (b) Pressure remains the same.
These short answers follow the rules of Boyle's Law directly. If you cut the volume in half (a), the molecules are squeezed, doubling the pressure, whereas if nothing changes (b), the pressure stays steady.
Teacher's Tip: This is a simple 2x relationship; half the space = double the hits.
Exam Tip: Always check if the question mentions that temperature is constant before applying this rule.

Question 10:
Answer: Charles's Law
At constant pressure, the volume of a given mass of a dry gas increases or decreases by 1/273 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): V₁ ∝ T₁ or V₁/T₁ = K = {Constant}
At temperature T₂ (K), Volume is V₂ (cm³): V₂ ∝ T₂ or {V₂}/{T₂} = K = {Constant}
therefore {V₁}/{T₁} = {V₂}/{T₂} = {Constant}
For Temperature = Conversion from Celsius to Kelvin
1{ K} = °C + 273
For example, 20°C = 20 + 273 = 293{ K}
Graphical representation of Charles 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.
Significance of Charles’ Law: Since the volume of a given mass of gas is directly proportional to its temperature, hence the density decreases with temperature. This is the reason that:
(a) Hot air is filled in the balloons used for meteorological purposes. (b) Cable wires contract in winters and expand in summers.
Charles's Law explains why things "grow" when they get hot and "shrink" when they get cold. This is because heat gives molecules the energy to push outward and claim more territory for themselves.
Teacher's Tip: Hot air is "lighter" (less dense) than cold air, which is why hot air balloons can fly!
Exam Tip: When using Charles's Law, the temperature must ALWAYS be in Kelvin, never Celsius.

Question 11:
Answer: Explanation Of Charles’ Law on the basis of kinetic theory of matter is as follows:
According to kinetic theory of matter, the average kinetic energy of the 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’ law).
If the container is flexible, like a balloon, the faster molecules will push the walls out until the internal pressure matches the outside air again. This results in a larger volume, proving that heat literally "stretches" the gas.
Teacher's Tip: Think of the molecules like "battering rams" getting stronger and faster as they get hotter.
Exam Tip: Mention that molecules strike walls "more frequently and vigorously" to explain expansion.

Question 12:
Answer: Absolute zero
The temperature -273°C is called absolute zero.
V = V₀ ({273 + t}{273})
Volume at -273°C = V₀ ({273 - 273}{273}) = 0
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 into Kelvin scale by adding 273 to it. For example, 20°C = 20 + 273 = 293{ K}.
Absolute zero is the coldest possible temperature in the universe where all molecular motion stops. It is the "starting line" for the Kelvin scale, which is why you will never see a negative number in Kelvin.
Teacher's Tip: At Absolute Zero, molecules are "frozen" in time and don't move at all!
Exam Tip: Define Absolute Zero as the temperature where the volume of a gas theoretically becomes zero.

Question 13:
Answer: (a) 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 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
Kelvin scale can be converted to degree Celsius by subtracting 273 from it. So, boiling point of water on centigrade scale is : 373{ K} - 273 = 100°C.
The Kelvin scale is the preferred tool for scientists because it eliminates the confusion of working with negative temperatures. Since gas volume cannot be less than zero, it makes sense to have a temperature scale that also starts at zero.
Teacher's Tip: Adding 273 is like "shifting" the whole thermometer up to a safer starting point.
Exam Tip: Remember that the size of 1 degree Celsius is exactly the same as 1 Kelvin; only the starting point is different.

Question 14:
Answer: (a) Standard or Normal Temperature and Pressure (S.T.P. or N.T.P.)
The pressure of the atmosphere which is equal to 76 cm or 760 mm of mercury is referred to as S.T.P. or N.T.P. The full form for S.T.P. is Standard Temperature and Pressure or Normal Standard temperature and pressure denotes 0°C or 273K.
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.
Standard values are like a common "baseline" so scientists can compare experiments done in different cities or climates. Without STP, a volume of gas measured in the hot tropics would look very different from the same gas measured in a cold laboratory.
Teacher's Tip: STP is like the "Average Joe" conditions for gas measurements.
Exam Tip: Be sure to know both units for pressure at STP: 1 atm and 760 mm Hg.

Answer (cont.): (b) Because the volume of a given mass of dry enclosed gas depends upon the pressure of the gas and temperature of the gas in Kelvin so to express the volume of the gases we compare these to S.T.P.
Comparing gases to STP allows us to see the true nature of a substance without the local weather getting in the way. It turns a variable measurement into a standard fact that anyone in the world can understand.
Teacher's Tip: It's like measuring the height of a person; it's better to do it without their shoes on so everyone is compared fairly!
Exam Tip: Mention that gas volume is "highly sensitive" to pressure and temperature changes.

Question 15:
Answer: (a) (i) C = °C (ii) K = 273{K}
(b) (i) 1{ atm} (ii) 760{ mm Hg} (iii) 76{ cm Hg.} (iv) 1{ torr} = 133.32{ Pascal}
These are the standard conversion factors used in every gas law calculation. Knowing that 1 atm is equal to 760 mm of mercury allows you to switch between units depending on what the question provides.
Teacher's Tip: 1{ torr} and 1{ mm Hg} are actually the exact same amount of pressure!
Exam Tip: Memorize the value of 1 Torr in Pascals (133.32{ Pa}) for advanced physics questions.

Question 16:
Answer: Temperature on Kelvin scale (K) = 273 + Temperature on 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
These calculations show how to "hop" between the two main temperature scales used in school. Adding 273 moves you from the freezing-point scale to the absolute-energy scale.
Teacher's Tip: To go from Celsius to Kelvin, you always add. To go back, you always subtract!
Exam Tip: Never put a degree symbol (°) next to the letter 'K' for Kelvin; it is just 'K'.

Question 17:
Answer: (a) Charles’s Law (b) Boyles Law
Boyle's law focuses on the relationship between pressure and volume, while Charles's Law connects temperature and volume. Together with Gay-Lussac's Law, they form the foundation of gas physics.
Teacher's Tip: Think: "Boyle" is under "Pressure" (P-V), and "Charles" is "Hot" (T-V).
Exam Tip: Identifying which law to use is the first step to solving any numerical problem.

Question 18:
Answer: (a) The real advantage of the Kelvin scale is that it makes the application and use of gas laws simple. Even more significantly, all values on the scale are positive. Thus, removing the problem of negative (-) values on the Celsius scale.
(b) The mass of a gas per unit volume is very small due to the large intermolecular spaces between the molecules. Therefore, gases have low density. Whereas 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) Since the volume of a gas changes remarkably with change in temperature and pressure, it becomes necessary to choose standard value of temperature pressure.
(e) According to Boyle’s Law, the volume of a given mass of a gas is inversely proportional to its pressure at constant temperature.
V ∝ 1/P
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 balloon, the volume of air also decreases. And this violates Boyle’s law.
(f) Atmospheric pressure is very low at high altitudes, volume of air increases thus 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) In gas as 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 and takes the shape of the vessel in which it is kept.
These qualitative reasons explain why the physical laws we study actually matter in real life, like mountain climbing or inflating a balloon. Gases are unique because their density and volume are so flexible compared to the rigid nature of solids.
Teacher's Tip: For part (e), note that the mass of gas in an inflating balloon is NOT constant, which is why Boyle's Law seems to fail there.
Exam Tip: Use the term "intermolecular space" when explaining why gases have such low density.

Question 19:
Answer: The temperature scale with its zero at -273°C and where each degree is equal to degree on the Celsius scale is called the absolute scale of temperature.
The temperature -273°C is called the absolute zero. Theoretically, this is the lowest temperature that can never be reached. At this temperature all molecular motion ceases.
The temperature - 273°C is called absolute zero.
V = V₀ {273 + t}/{273})
Volume at -273°C = V₀ {273 - 273}/{273}) = 0
Scientists use the idea of absolute zero to define a perfect starting point for energy. If a gas could actually reach this temperature, it would lose all its energy and technically take up no space at all.
Teacher's Tip: Absolute zero is the "Ultimate Cold" - nothing in the universe can be colder!
Exam Tip: Remember the specific number -273°C; it is the most common answer for "What is absolute zero?"

Question 20:
Answer: Gases like nitrogen, hydrogen are collected over water as shown in the figure. 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.
P_total = P_gas + P_{water vapour}
P_gas = P_gas - P_{water vapour}
Actual Pressure of gas = Total pressure – Aqueous tension
When you collect a gas in a lab using water, the gas gets "wet" and picks up a tiny bit of water pressure. To find the true pressure of just the gas, you have to subtract this "aqueous tension" like taking the weight of a box away from the weight of the items inside.
Teacher's Tip: Think of "Aqueous Tension" as the "tax" you pay for using water to collect your gas.
Exam Tip: Always subtract the aqueous tension from the given total pressure before starting your numerical calculation.

Question 21:
Answer: (a) The volume of gas is zero. (b) The absolute temperature is 7 + 273 = 280{ K} (c) The gas equation is:
P₁V₁ /{T₁} = {P₂V₂}/T₂
(d) Ice point = 0 + 273 = 273{ K} (e) Standard Temperature is taken as 273{ K} or 0°C. Standard pressure is taken as 1 atmosphere (atm) or 760 mm Hg.
The "Combined Gas Law" equation in part (c) is the most powerful tool because it lets you solve problems where pressure, volume, AND temperature are all changing at once. It combines the rules of Boyle and Charles into one easy formula.
Teacher's Tip: This master formula is like the "Swiss Army Knife" of gas laws.
Exam Tip: Practice rearranging the gas equation to solve for V₂ or P₂ before the exam starts.

Question 22:
Answer: (a) (iii) Straight line parallel to X- axis. (b) (ii) 27°C = 27 + 273 = 300{ K} (c) (iv) Charles (d) (ii) 1/2 times
These multiple-choice options highlight key relationships, such as how the product of pressure and volume (PV) stays perfectly flat on a graph if the temperature doesn't move. It also reminds us that doubling the pressure must result in the volume being cut in half (1/2).
Teacher's Tip: For part (d), remember: Inverse = Upside down. If Pressure is 2, Volume is 1/2.
Exam Tip: Don't rush these! A simple addition error in a temperature conversion (27 + 273) is the most common way to lose points.

Question 23:
Answer:
Column A | Column B
(a) Cm^{3} | Volume
(b) Kelvin | Temperature
(c) Torr | Pressure
(d) Boyles law | PV = P₁V₁
(e) Charles's law | V/T = V₁/T₁
This table links the scientific units we use in the lab with the variables they measure. For example, Torr is a specialized unit for pressure named after the scientist Torricelli, who invented the first barometer.
Teacher's Tip: Cm³ is the same as 1{ Milliliter} (ml).
Exam Tip: Make sure you know which equation belongs to which law for matching questions.

Question 24:
Answer: (a) Volume of a gas is directly proportional to the pressure at constant temperature. (False: Inversely)
(b) Volume of a fixed mass of a gas is inversely proportional to the temperature, the pressure remaining constant. (False: Directly)
(c) -273°C is equal to zero Kelvin. (True)
(d) Standard temperature is 0°C (True)
(e) The boiling point of water is 373 K. (True)
This exercise corrects common misconceptions about gas behavior. It is vital to remember that temperature and volume go up together (direct), but pressure and volume go in opposite directions (inverse).
Teacher's Tip: Re-read "directly" and "inversely" twice - they are the most common trick words!
Exam Tip: If a question says something is "False," always provide the corrected statement to show you know WHY it is wrong.

Question 25:
Answer: (a) Absolute temperature (b) Absolute zero (c) Volume (d) 273
These four terms are the core "vocabulary" of this chapter. Mastering them allows you to follow the logic of every gas law experiment and calculation.
Teacher's Tip: "Absolute" always refers to the Kelvin scale!
Exam Tip: Fill in blanks carefully; one wrong word can change the entire physical law.

Numericals

Numerical Solution 1:
Answer: Given: V = 800 { cm}³, P = 650 { m}, P₁ = ?
V₁ = {reduced volume} = 40% { of } 800
= 800 × {40}/{100} = 320
Net V₁= 800 - 320 = 480 { cm}³
T = T₁
Using gas equation, we get,
{PV}/{T} = {P₁V₁}/{T₁}
{800 × 650}/{T} = {P₁ × 480}/T
Since, T = T₁
therefore P₁ = {800 × 650}/{480}
= 1083.33 { mm of Hg.}
This problem tests your ability to handle percentages. When a volume is "reduced by 40%," it means you have 60% of the original space left to calculate with.
Teacher's Tip: Always calculate the *final* volume (V₁) before plugging numbers into the equation.
Exam Tip: If the temperatures are the same, you can simply cross them out of both sides of the equation.

Numerical Solution 2:
Answer: Let, V₁ = x, V₂ = ?
P₁ = 1 { atm.} P₂ = 2 { atm.}
T₁, T₂ = 3T₁
{P₁V₁}/{T₁} = {P₂V₂}/{T₂}
{1 × x}{T₁} = {2 × V₂}{3T₁}
V₂ = {3T₁ × x}/{T₁× 2}
V₂ = 1{1}/{2} { times original volume } V₁
Even though the pressure doubled (which should shrink the gas), the temperature tripled (which should expand it). Because the temperature increase was stronger than the pressure increase, the gas ultimately ended up 1.5 times larger.
Teacher's Tip: When you don't have numbers, use "x" to represent the original amount.
Exam Tip: State your final answer as a "multiplier" of the original volume (like 1.5 times) to be clear.

Numerical Solution 3:
Answer: P = 100, V = 20
P₁ = 1, V = ?
T = T₁
Using equation,
{PV}/{T} = {P₁V₁}/{T₁}
{100 × 20}/{T} = {1 × V₁}/{T}
therefore V₁ = 2000 { lit.} = 2 m³ (Since, 1000 { lit.} = 1 m³)
Volume of one flask = 200 { cm}³ = {200}/{100 × 100 × 100} m³
therefore {Number of flasks} = {2 × 1000000}/200
= 10000
Number of flasks = 10000
This is a practical problem about industrial storage. It shows how much gas you can fit into small containers once you release it from a high-pressure tank into normal room conditions.
Teacher's Tip: Be very careful with unit conversions between Liters, m³, and cm³!
Exam Tip: Always show the conversion (1000 { lit} = 1 m³) so the examiner knows how you got your final numbers.

Numerical Solution 4:
Answer: V = 20 { lit.} P = 29 { atm}
P₁ = 1.25 { atm } V₁ = ?
t = t₁
Using gas equation,
{PV}/{t} = {P₁V₁}/{t₁}
{29 × 20}/{t} = {1.25 × V₁}/{t}
therefore V₁ = {29 × 20}/{1.25}
V₁ = 464 { liters}
By reducing the pressure from 29 atm down to just 1.25 atm, the gas expanded significantly. The final volume of 464 liters is more than 23 times the size of the original 20-liter container.
Teacher's Tip: If the pressure goes down, your final answer for volume MUST be larger than the starting volume.
Exam Tip: Use the "PV = Constant" shortcut if the temperature is not mentioned as changing.

Numerical Solution 6:
Answer: V = 3 { liters } t = 0°C = 0 + 273 = 273 { K}
V₁= ?
T₁ = -20°C = -20 + 273 = 253 { K}
P = P₁
Using gas equation,
{PV}/{T} = {P₁V₁}/{T₁}
{3 × P}/{273} = {V₁ × P}/{253} [because P = P₁]
therefore V₁ = {3 × 253}/{273}
= 2.78 { liters}
therefore V₁ = 2.78 { liters}
This problem follows Charles's Law because the pressure is constant. When the temperature dropped from 0°C to -20°C, the gas contracted, reducing the volume from 3 liters to 2.78 liters.
Teacher's Tip: Cold temperatures always make gases "huddle closer together," reducing their volume.
Exam Tip: Even for simple ratios, write out the formula {V₁}/{T₁} = {V₂}/{T₂} to get your method marks.

Numerical Solution 7:
Answer: (a) V = 2 { liters } P = 760 { mm}
V₁ = 4 { Dm}³ = 4 × 10 {m} × 10 {m} × 10 {m} = 4000 { m}³} [since, 1 { Dm}^{3} = 1000 { m}³]
= 4 { lit.} [{Since, } 1 { lit} = 1000 { m}³]
P₁ = ?
t = t₁
Using gas equation,
{PV}/{t} = {P₁V₁}/{t₁}
{2 × 760}/{t} = {P₁ × 4}/{t₁} [because t = t₁]
P₁ = {2 × 760}/{4} = {760}/{2} = 380
P₁ = 380 { mm.}
In this calculation, we see that doubling the volume (2 liters to 4 liters) resulted in the pressure being cut exactly in half (760 mm to 380 mm). This is a perfect real-world example of Boyle's inverse relationship.
Teacher's Tip: Volume up = Pressure down. It's a simple inverse rule!
Exam Tip: Be extremely careful with units like Dm³; always convert them to liters first to avoid mistakes.

Numerical Solution 8:
Answer: V = 500 { cm}³
Normal temperature, t = 0°C = 0 + 273 = 273 { K}
V₁= {Reduced volume} + 20% { of } 500 { cm}³
= {20 × 500}/{100} = 100 { cm}³
Net, V₁ = 500 - 100 = 400 { cm}³
T₁ = ?
P = P₁
Using gas equation, we get,
{PV}/{t} = {P₁V₁}/{t₁}
{P × 500}/{273} = {P × 400}/{t₁}
t₁ = {273 × 4}/{5} = 218.4 { K}
therefore t₁ = 218.4 - 273
= -52.60°C
To shrink the gas by 20%, we had to significantly lower the temperature. The calculation shows the gas had to be cooled from the freezing point (0°C) all the way down to -52.6°C.
Teacher's Tip: Don't forget that 20% is just the "cut," you must use the remaining 80% (400 { cm}³) as your final volume.
Exam Tip: Always convert your final Kelvin result back into Celsius if the question started with Celsius.

Numerical Solution 9:
Answer: V = 30 { cm}³
t = 15°C = 15 + 273 = 288 { K } P = 740 { mm}
At S.T.P.
V₁ = ?
t₁ = 0°C = 273 { K } P₁ = 760 { mm}
Using gas equation, {PV}/{t} = {P₁V₁}/{t₁}
{30 × 740}/{288} = {760 × V₁}/{273}
V₁ = {30 × 740 × 273}/{288 × 760} = 27.68 = 27.7 { cm}³
therefore V₁ = 27.7 { cm}³
This problem uses the Combined Gas Law to adjust a lab measurement to Standard Temperature and Pressure. Even though both temperature and pressure changed slightly, the combined formula gives us the precise volume the gas would occupy at STP.
Teacher's Tip: Rounding 27.68 to 27.7 is standard practice in chemistry labs for significant figures.
Exam Tip: Clearly list your "Initial Conditions" and "Final Conditions" (STP) before starting the math.

Numerical Solution 10:
Answer: P = 770 { mm } V = 88 { cm}³ P₁ = 880 { mm } V₁= ?
t = t₁
Using gas equation,
{PV}/{t} = {P₁V₁}/t₁
Or 770 × 88 = V₁ × 880
therefore V₁ = 77 { cm}³
therefore {Volume diminished} = 88 - 77 = 11 { cm}³
By increasing the pressure from 770 to 880, we pushed the molecules closer together, shrinking the volume. The "diminished" volume is the amount of space that was lost during this compression.
Teacher's Tip: "Diminished" just means "subtracted." Find the difference between start and finish.
Exam Tip: Read carefully: if asked for the "diminished volume," the answer is 11, not 77.

Numerical Solution 11:
Answer: Pressure of dry hydrogen P = 750 - 14 = 736 { mm } V = 50 { cm}³
t = 17°C = 17 + 273 = 290 { K}
P₁= 760 { mm}
V = ?
T₁ = 0°C = 273 { K}
Using gas equation,
{PV}/{t} = {P₁V₁}/{t₁}
{736 × 50}/{290} = {760 × V₁}/{273}
therefore V₁ = {736 × 50 × 273}/{290 × 760}
= 45.85 { cm}³
Let us say (By rounding up),
= 45.6 { cm}³
therefore V₁= 45.6 { cm}³
This is a complex problem because it accounts for "moist" gas by subtracting aqueous tension first. Only after you have the pressure of the "dry" gas can you use the Combined Gas Law to find the volume at STP.
Teacher's Tip: Dry Pressure = Total Pressure - Water Pressure. Always do this step first!
Exam Tip: Rounding to one decimal place (45.6) is usually expected unless the question asks for more precision.

Numerical Solution 12:
Answer: t = C = 273 { K } P = 760 { mm}
V = 100 { cm}³
t₁ = 273 × {1}/{5} = 54.6 { (Increased)} = 273 + 54.6 = 327.6 { K}
P₁ = 1{1}/{2} × 760 = 760 × {3}/{2} = 1140 { mm}
V₁ = ?
Using gas equation,
{PV}{t} = {P₁V₁/{t₁} = {760 × 100}/273 = {1140 × V₁ /{327.6}
therefore V₁ = {760 × 100 × 327.6}/{273 × 1140} = 80 { cm}³
therefore V₁ = 80 { cm}³
In this case, the pressure was increased by 50%, which tried to shrink the gas, while the temperature also increased, trying to expand it. The pressure's effect was stronger, so the gas ended up smaller (80 { cm}³) than its starting size.
Teacher's Tip: "Increased by 1/5" means you add 20% to the original Kelvin value.
Exam Tip: Convert "one and a half times" into the fraction 3/2 immediately to make the math easier.

Numerical Solution 13:
Answer: Let original volume (V) = 1 and original pressure (P) = 1 and temp given (t) = -15°C = -15 + 273 = 258 { K}
V₁ or New volume after heating = original volume + 50% of original volume
= 1 + {50}/{100} = 1 + {1}/{2} = {3}/{2}
P₁ or decreased Pressure = 60%
= 1 × {60}/{100} = .6
t₁ = {To be calculated.}
because {PV}/{t} = {P₁V₁ /{t₁} → {1 × 1}/{258} = {.6 × {3}/{2}}{t₁} = {.9}/{t₁}
therefore t₁ = {.9 × 258}{1} = 232.2 { K}
therefore t₁ = 232.2 - 273 = -40.8°C.
This problem shows how temperature must change to accommodate shifts in both volume and pressure. By setting the starting values to '1', we can easily see how the percentages change the whole system.
Teacher's Tip: Using "1" as a starting point is a great trick for any percentage-based gas law problem.
Exam Tip: Be very careful with the negative sign when converting 232.2 K back to Celsius! (-40.8°C).

Numerical Solution 14:
Answer: V = 2 { lit.}
P = 100 { pascal } t = 27°C = 273 + 27 = 300 { K}
P₁ = {1}/{2} { of } 100 = {1}/{2} × 100 = 50 { Pascal}
V₁ = {1}/{2} { of } V = {1}/{2} × 2 = 1 { lit.}
t₁ = ?
Using gas equation,
{PV}/{t} = {P₁V₁}/{t₁} → {100 × 2}/{300} = {50 × 1}/{t₁}
therefore t₁= {50 × 300}/{100 × 2}
therefore t₁ = 25 × 3 = 75 { K}
therefore t₁ = 75 - 273
t₁ = -198°C
If you want to cut both the pressure AND the volume in half, you have to cool the gas down immensely. This calculation shows the gas would need to drop from room temperature all the way down to -198°C.
Teacher's Tip: This is a massive temperature drop, close to the temperature of liquid nitrogen!
Exam Tip: Cross-multiply carefully to avoid mixing up your P₁ and V₁ in the final step.

Numerical Solution 15:
Answer: (Initial Conditions) (Final Conditions)
P₁= 1 { atm } = 760 { mm } P₂ = 760 × {5}/{2} + 760 = 2660 { mm}
V₁ = 2500 { cm}³ V₂ = ?
T₁ = 273 + 0 = 273 { K } T₂ = 273 { K}
Applying general gas equation,
{P₁V₁/{T₁} = {P₂V₂}/{T₂} → {760 × 2500}/{273} = {2660 × V₂}/{273}
V₂ = {760 × 2500}/{2660} × {273}/{273} = {5000}/{7} { cc} = 714{2}/{7} { cc}
= 5000/7 { cm}³
This problem calculates how much a volume of gas shrinks when the pressure is massively increased. Since the temperature didn't change, the gas was simply compressed into a much tighter space of about 714 cubic centimeters.
Teacher's Tip: "cc" and "cm³" are the exact same thing; don't let the different letters confuse you!
Exam Tip: Fractions like 5000/7 are sometimes better than decimals in chemistry to avoid rounding errors too early.

Numerical Solution 16:
Answer: (Initial Conditions) (Final Conditions)
P₁ = P { mm } = 760 { mm } P₂ = 760 × {5}/{2} + 760 = 2660 { mm}
V₁ = 5000/7 { cc } V₂= 2500 { cc}
T₁ = 273 { K } T₂ = ?
Applying general gas equation,
{P₁V₁/{T₁} = {P₂V₂}/{₂} → {P × 5000}/{273 × 7} = {P × 2500}/{T₂}
T₂ = {P × 2500 × 273 × 7}/{P × 5000} = {7}/{2} × 273 { K} = 3.5 { Times}
By solving this, we find that to expand the gas back to 2500 cc while under high pressure, we would need to increase the absolute temperature by 3.5 times. This illustrates the massive energy required to combat high pressure in a gas system.
Teacher's Tip: Notice how the unknown pressure 'P' cancels out entirely, making the calculation much simpler.
Exam Tip: Use "3.5 Times" as part of your answer to describe the relationship clearly.

Numerical Solution 17:
Answer: (a) Initial Condition / Final Condition
V₁ = V {cc } V₂ = V{cc}
P₁ = 100 { cm of Hg } P₂ = 10 { cm of Hg}
T₁ = 0°C = 273 { K } T₂ = ?
Applying general gas equation,
{P₁V₁}/{T₁} = {P₂V₂}/{T₂}
{100 × V}/{273} = {10 × V}/{T₂}
T₂ = {273 × 10}/{100}
T₂ = 27.3 { K}
(b) Initial Conditions / Final Conditions
V₁ = V {cc } V₂ = V{cc}
P₁ = 100 { cm of Hg } P₂ = ?
T₁ = 0 + 273 = 273 { K } T₂ = 273 + 100 = 373 { K}
Applying general gas equation,
{P₁V₁ /{T₁} = {P₂V₂}/{T₂}
{100 × V}/{273} = {P₂ × V}/{373}
P₂ = {373 × 100}/{273}
P₂= 136.63 { cm of Hg}
These two parts show two different scenarios: in (a) we see how cold a gas must get to lower its pressure, and in (b) we see how much the pressure climbs just by heating the gas up by 100 degrees.
Teacher's Tip: Since volume stays the same (V), this is technically an application of Gay-Lussac's Law.
Exam Tip: Be sure to label each part of the problem (a and b) clearly to avoid losing points.

Numerical Solution 18:
Answer: Capacity of the cylinder V = 10000 { lit.}
P = 800 { mm}
t = -3°C = -3 + 273 = 270 { K } P₁ = 400 { mm of Hg}
t₁ = 0°C = 0 + 273 = 273 { K}
V₁= ?
Using gas equation,
{PV}/{t} = {P₁V₁}/{t₁}
therefore {800 × 10000}/{270} = {400 × V₁}/{273}
therefore V₁= {800 × 10000 × 273}/{270 × 400}
= 20222.2 { lit.}
Number of Cylinders = V₁/Volume of one cylinder
= {20222.2}/{10} = 2022.22
= 2022 { Cylinders}
This is a great engineering problem! It shows how to calculate the total number of portable tanks you can fill from one giant storage tank by adjusting for changes in pressure and temperature.
Teacher's Tip: Always round down for the number of full cylinders, because you can't have 0.22 of a tank!
Exam Tip: The units for V and V₁ must be the same (liters) for the division to work.

Numerical Solution 19:
Answer: Volume of 1 mole of gas at STP = 22.4 { lit.}
V₁ = 22.4 { liters } V₂ = ?
T₁ = 273 { K } T₂= 27 + 273 = 300 { K}
P₁= 1 { atm } P₂ = 4 { atm}
Using gas equation,
{P₁V₁}/{T₁} = {P₂V₂}/{T₂}
therefore V₂ = {P₁V₁T₂}/{T₁P₂}
= {1 × 22.4 × 300}/{273 × 4}
= {560}/{91}
V₂ = 6.15 { Liters}
This numerical uses the "Molar Volume" constant (22.4{ L}) to find how much space one mole of gas takes up in high-pressure laboratory conditions. Even though it was heated, the extra pressure squashed it down to less than a third of its original size.
Teacher's Tip: Remember 22.4 liters as the "Magic Number" for one mole of gas at STP.
Exam Tip: Be very careful when plugging five different numbers into one formula! Double-check your setup.

Numerical Solution 20:
Answer: Applying equation,
{PV}/{t} = {P₁V₁}/t₁
P₁= 760 { mm } P₂} = 50% { of } P₁ = .50 × 760 = 380 { mm}
V₁ = V₁ V₂ = 2V₁
t₁= 273 { K } t₂ = ?
therefore {760 × V₁}/{273} = {380 × 2V₁}/{t₂}
therefore t₂ = {380 × 2₁}/{760 × V₁
therefore t₂ = 273 { K}
This unique problem shows that if you double the volume but cut the pressure in half, they perfectly cancel each other out. Because the "VP" product didn't change, the temperature also remained exactly the same as the starting temperature.
Teacher's Tip: This is a "trick" problem where the final answer matches the starting answer!
Exam Tip: Show that V₁ cancels out on both sides to prove your mathematical logic.

Numerical Solution 21:
Answer: (a) P = 748 { mm Hg } V = 1.2 {l}
T = 25°C = 25 + 273 = 298 { K}
P₁ = 760 { mm of Hg}
V₁ = ?
T₁ = 273 { K}
Applying general gas equation,
{PV}/{T} = {P₁V₁}/{T₁}
therefore {748 × 1.2}/{298} = {760 × V₁}/{273}
therefore V₁ = {748 × 1.2 × 273}/{298 × 760} = 1.081 { liters}
(b) P = 760 { mm Hg } P₁ = 760 { mm Hg } V = 1.25 { / } V₁ = ?
T = 273 { K } T₁ = 273 { K}
Applying general gas equation,
{PV}/{T} = {P₁V₁}/{T₁} → {760 × 1.25}/{273} = {760 × V₁}/{273} = 1.25 { l}
1.25{ l } O₂ at S.T.P. will have greater volume than 1.2 { l } N₂ at 25°C and 748 mm Hg, when the 2 gases are compared at S.T.P.
Comparing two different gases requires bringing both of them to the "neutral ground" of STP first. Only after converting the Nitrogen from lab conditions to STP conditions (1.081{ L}) can we see that the Oxygen sample was actually larger.
Teacher's Tip: You can only compare apples to apples when they are at the same temperature and pressure!
Exam Tip: Be sure to write a final concluding sentence that answers the "which is greater" question.

Numerical Solution 22:
Answer: P₁ = 1 { bar } P₂ = ?
V₁ = 500 { dm}³ V₂ = 200 { dm}³
T₁ = 273 { K } T₂ = 273 + 30 = 303 { K}
Using gas equation,
{P₁V₁}/{T₁} = {P₂V₂}/{T₂}
P₂ = {P₁V₁T₂}/{T₁V₂}
= {1 × 500 × 303}/{273 × 200} = {5 × 101}/{91} = {505}/{91}
therefore P₂ = 2.77 { bar}
In this problem, the pressure increased from 1 bar to 2.77 bar because we squeezed the gas into a much smaller tank and heated it up at the same time. Both of these changes work together to drive the internal pressure higher.
Teacher's Tip: "Bar" is another common unit for pressure, just like Pascals or Atmospheres.
Exam Tip: Check your final result: smaller volume + higher heat should definitely equal higher pressure.

Numerical Solution 23:
Answer: (i) V₁ = 0.4 { L } V₂ = 0.4 × 2 { L}
T₁ = 17°C (17 + 273) = 290 { K } T₂= ?
{V₁}/{T₁} = {V₂}/{T₂}
{0.4}/{290} = {0.8}/{T₂}
therefore T₂ = {290 × 0.8}/{0.4} = 290 × 2 = 580
= 580 - 273 = 307 { K}
(ii) V₁ = 0.4 { L } V₂ = 0.2 {L}
T₁ = 17°C (17 + 273) = 290 { K } T₂ = ?
{V₁}/{T₁} = {V₂}/{T₂}
{0.4}/{290} = {0.2}/{T₂}
therefore T₂ = {290 × 0.2}/{0.4} = 145 { K}
= 145 - 273 = -128°C
These two examples show the "Volume-Temperature" link in Charles's Law: to double the volume, you must double the absolute temperature; to halve the volume, you must cut the absolute temperature in half.
Teacher's Tip: If the volume doubles, the Kelvin temperature MUST also double. It's a perfect 1-to-1 ratio!
Exam Tip: Always subtract 273 to turn your final Kelvin answer back into Celsius if needed.

Numerical Solution 24:
Answer: Pressure due to dry air, P = 750 - 12 = 738 { mm } V = 28 { cm}³
t = 14°C = 14 + 273 = 287 { K}
P₁ = 760 { mm of Hg } V₁ = ?
t₁ = 0°C = 273 { K}
Using gas equation,
{PV}/{t} = {P₁V₁}/{t₁}
{738 × 28}/{287} = {V₁ × 760}/{273}
therefore V₁ = {738 × 28 × 273}/{287 × 760}
= 25.86 { cm}³ = 25.9 { cm}³
therefore V₁= 25.9 { cm}³
This is another practical laboratory calculation where we clean the measurement by removing aqueous tension and then shifting it to STP. The final answer (25.9 cubic centimeters) is the "true" standard volume of that specific sample of gas.
Teacher's Tip: Aqueous tension is basically "humidity" that adds extra pressure to your sample.
Exam Tip: Show the subtraction (750 - 12) as its own clear step at the very top of your work.

Numerical Solution 25:
Answer: Step 1: To convert the volume of the gas to S.T.P.
Initial conditions / Final Conditions
P₁= 700 { mm } P₂ = 760 { mm}
V₁ = 20 { L } V₂= ?
T₁ = 273 + 27 = 300 { K } T₂ = 273 { K}
Applying general gas equation,
{P₁V₁}/{T₁} = {P₂V₂}/{T₂} → {700 × 20}/{300} = {760 × V₂}/{273}
therefore V₂ = {273 × 700 × 20}/{760 × 300} = {91 × 14}/{38 × 3} = {91 × 7}/{38 × 3} = 637/38
Step 2: To calculate weight of the gas.
22.4 { liters of the gas weighs at S.T.P.} = {70 × 637}/{22.4 × 38}
= {700 × 637 × 100 × 637}/{224 × 38 × 32 × 38}
= {25 × 637}/{8 × 38}
= 52.38 { g.}
This final numerical combines gas laws with the concept of density and molar mass. By finding the volume at STP first, we can use the standard molar volume of 22.4 liters to find the actual physical weight of the gas in grams.
Teacher's Tip: This is a two-level boss fight! Finish Step 1 completely before you even look at Step 2.
Exam Tip: Be very accurate with your division in Step 2; a small error here will throw off the final weight measurement.