Frank Brothers Solutions for ICSE Class 9 Chemistry Chapter 2 Study Of Gas Laws

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Question 1:
Answer: An ideal gas can be characterized by three state variables:
1. Absolute pressure (P),
2. Volume (V), and
3. Absolute temperature (T).
These variables are measurable properties that define the physical state of a gas sample. By knowing these three factors, scientists can predict how a gas will behave when its environment changes.
Teacher's Tip: Remember the "PVT" system - Pressure, Volume, and Temperature are the big three of gas laws.
Exam Tip: Always capitalize the symbols P, V, and T when representing these state variables in equations.

Question 2:
Answer: Boyle’s law states that “At a constant temperature the volume of a fixed mass of gas is inversely proportional to its pressure.”
V ∝ 1/P (At constant temperature)
This means that if you squeeze a gas into a smaller space, the pressure inside that space will increase. If you increase the volume by giving it more room, the pressure will drop accordingly.
Teacher's Tip: Think of a bicycle pump; when you push the handle down, the volume gets smaller and the pressure gets much higher.
Exam Tip: Don't forget to mention the condition "at constant temperature" as the law is invalid without it.

Question 3:
Answer: At constant temperature, volume of a given mass of a gas is inversely proportional to its pressure.
V ∝ 1/P
therefore V = K/P
Where, K is the constant of proportionality.
This mathematical expression allows us to calculate changes in a gas sample. Since V × P = K, we know that the product of pressure and volume remains the same as long as temperature doesn't change.
Teacher's Tip: If pressure doubles, volume must become half to keep the "K" constant.
Exam Tip: Define "K" as the proportionality constant to show a complete mathematical understanding.

Question 4:
Answer: Charles’ law states that “At constant pressure, the volume of a given mass of a dry gas is directly proportional to its absolute temperature (Kelvin).”
V ∝ T (At constant pressure)
This law explains that gases expand when they are heated and shrink when they are cooled. It specifically requires the use of the Kelvin scale because it relates to the total energy of the gas particles.
Teacher's Tip: Think of a hot air balloon; the air inside is heated, it expands (increases volume), and becomes lighter than the air outside.
Exam Tip: Always specify "Absolute Temperature" or "Kelvin" because using Celsius in this formula will give the wrong answer.

Question 5:
Answer: Kelvin zero is - 273.15 ° C.
This point, also known as absolute zero, is the coldest possible temperature where all molecular motion theoretically stops. Scientists use this as the starting point for the Kelvin scale because there are no negative temperatures in Kelvin.
Teacher's Tip: To convert Celsius to Kelvin, just add 273 to your Celsius number.
Exam Tip: Note that Kelvin does not use the "degree" symbol; write it as "K", not "° K".

Question 6:
Answer: According to Boyle’s law,
V ∝ 1/P (At constant temperature)
or PV=K ...........(1)
According to Charles’ law
V ∝ T (At constant pressure)
or V/T = K ...........(2)
On combining the above laws,
PV/T = K ...........(3)
Where, K is a constant.
Let us suppose that, for a given mass of a gas, the initial pressure, volume and temperature are P₁, V₁ and T₁ which changes to P₂, V₂ and T₂ respectively. Then,
{P₁V₁}{T₁} = K and P₂V₂}{T₂} = K
Therefore,
{P₁V₁}{T₁} = {P₂V₂}/{T₂} ...........(4)
Equation (4) is called combined gas law equation.
The combined gas law is useful when pressure, volume, and temperature are all changing at the same time. It allows us to solve for any one missing variable if we know the other five initial and final values.
Teacher's Tip: Remember: Pressure and Volume are on top, and Temperature is always on the bottom.
Exam Tip: When deriving this, clearly label your equations (1, 2, 3) to help the examiner follow your logic.

Question 7:
Answer: The standard temperature and pressure (STP) by general convention are 0 ° C (273 K) and 1 atm (760 mm Hg).
STP provides a uniform set of conditions so that scientists can compare gas samples from different experiments. Since gas volume changes so much with T and P, these fixed values act as a baseline.
Teacher's Tip: Think of STP as the "Standard Setting" for all gas calculations.
Exam Tip: Be sure to write both the Celsius and Kelvin values for temperature to provide a complete answer.

Question 8:
Answer: 1. The value of standard temperature is (i) 0 ° C and (ii) 273 K
2. The value of standard pressure is (i) 1 atm, (ii) 760 mm of Hg, (iii) 76 cm of Hg, (iv) 760 torr
Pressure can be measured in many different units depending on the instrument used. Knowing that 1 atmosphere is equal to 760 millimeters of mercury helps us convert between different types of weather and lab data.
Teacher's Tip: Torricelli (torr) and mm Hg are actually the same value! 1 torr = 1 mm Hg.
Exam Tip: Pay attention to units in word problems; you may need to convert cm of Hg to mm of Hg before calculating.

Question 9:
Answer: P₁ = 240{ mm Hg}, P₂ = 720{ mm Hg}
V₁ = 45.6cm³, V₂ = ?
According to Boyle’s law, at constant temperature,
P₁V₁ = P₂V₂
240 times 45.6 = 720 times V₂
V₂ = {240 times 45.6}/{720} = {10944}/{720} = 15.2{ cm}³
In this problem, the pressure increased significantly, so we expected the volume to shrink. The result shows that the gas was compressed into a space about three times smaller than where it started.
Teacher's Tip: Always check your answer logically; if pressure went up (from 240 to 720), the final volume MUST be smaller than the start.
Exam Tip: Show every step of the cross-multiplication to ensure you get partial marks even if you make a small calculation error.

Question 10:
Answer: (i) Graph of V vs P showing a downward curve.
(ii) Graph of V vs 1/P showing a straight line starting from the origin.
(iii) Graph of PV vs P showing a horizontal straight line.
Graphs are visual ways to prove gas laws. For example, a straight line in the V vs 1/P graph proves that volume and pressure have a perfect mathematical relationship.
Teacher's Tip: A straight line in science usually means two things are directly linked to each other.
Exam Tip: Label your X and Y axes clearly with units to avoid losing marks on graph questions.

Question 11:
Answer: Number of moles of ammonia gas = {6}/{22.4} = 0.26{ moles}
3 moles of hydrogen form = 2 mole of ammonia
0.26 moles of hydrogen form = {2 times 0.26}/{3} = 0.17{ moles of ammonia}
Volume of ammonia gas formed = 0.17 times 22.4 = 3.8{ litres}
This calculation uses the concept that one mole of any gas at STP occupies 22.4 litres. It helps us determine exactly how much product we will get from a chemical reaction involving gases.
Teacher's Tip: 22.4 is the "Magic Number" for gases at STP.
Exam Tip: Write down the molar ratio (3:2 in this case) clearly before starting your division.

Question 12:
Answer: There is simultaneous effect of temperature and pressure changes on the volume of a given mass of a gas. So, when stating the volume of a gas, the pressure and temperature should also be given.
Since volume is so sensitive to its environment, just saying "10 litres of gas" is not specific enough for scientists. You must provide the "context" of T and P so others can replicate your results.
Teacher's Tip: Think of it like a recipe; you can't just say "cook the cake," you have to say at what temperature and for how long.
Exam Tip: This is a common "Give Reason" question; use the phrase "simultaneous effect" in your answer.

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Question 13:
Answer: Yes, it is possible to change the temperature and pressure of a fixed mass of a gas without changing its volume. If the pressure and temperature of a given mass of a gas are changed simultaneously such that ratio of T and P remains constant, then volume will remain unchanged.
PV = nRT
V = {nRT}/{P} (where R is constant)
By balancing the increase in temperature with a specific increase in pressure, you can keep the gas in the same sized container. This is a common practice in industrial engineering to prevent gas tanks from expanding or contracting.
Teacher's Tip: If T goes up, P must go up by the same ratio to keep V the same.
Exam Tip: Mention the Ideal Gas Equation (PV=nRT) to back up your explanation with theory.

Question 14:
Answer: 1. Volume of a gas would be reduced to zero at 0 K (-273 ° C). All temperatures on the Kelvin scale are positive, so Kelvin scale has been adopted for chemical calculation.
2. At absolute zero temperature, volume of a gas would be reduced to zero. Theoretically, this is the lowest temperature that can be reached. At this temperature all molecular motions cease. Thus, practically this temperature is impossible to attain because on cooling gases liquefy and Charles’ law is no more applicable.
3. According to combined gas law equation, there is simultaneous effect of temperature and pressure changes on the volume of a given mass of a gas. So, when stating the volume of a gas, the pressure and temperature should also be given.
Absolute zero is a theoretical limit where matter has no thermal energy left at all. In the real world, gases turn into liquids or solids long before they reach that point, which changes how they behave.
Teacher's Tip: Absolute zero is like the "Speed Limit" of coldness; you can't go any lower.
Exam Tip: Explain that Charles' Law stops working when a gas turns into a liquid.

Question 15:
Answer: Given:
V₁ = 4{ litres}; V₂ = ?
T₁ = 27 ° C = 273 + 27 = 300{ K}; T₂ = 150{ K}
According to Charles’ law, at constant pressure,
{V₁}/{T₁} = {V₂}/{T₂}
{4}/{300} = {V₂}/{150}
therefore V₂ = {4 times 150}/{300} = 2{ litres}
The temperature was cut in half (from 300K to 150K), so the volume also became half (from 4L to 2L). This perfectly demonstrates the "direct proportionality" mentioned in Charles' Law.
Teacher's Tip: Always convert Celsius to Kelvin first! Most students fail this question because they use "27" instead of "300".
Exam Tip: Clearly show the addition of 273 in your given data section to get full marks for steps.

Question 16:
Answer: P₁ = 12{ atm}; P₂ = 14.9{ atm}
T₁ = 27 ° C = 273 + 27 = 300{ K}; T₂ = ?
{P₁}/{₁} = {P₂}/{T₂}
{12}/{300} = {14.9}/{T₂}
T₂ = {14.9 times 300}/{12} = 372.5{ K} = 372.5 - 273 = 99.5 ° C
As the gas is heated, its internal pressure increases because the particles are hitting the walls of the container much harder. To find the final temperature in Celsius, we simply subtract 273 from our final Kelvin result.
Teacher's Tip: Pressure and Temperature go up together; this is sometimes called Gay-Lussac's Law.
Exam Tip: If the question starts with Celsius, always provide the final answer back in Celsius unless asked otherwise.

Question 17:
Answer: Given;
P₁ = 760{ mm Hg}; P₂ = {3}/{2} times 760 = 1140{ mm Hg}
V₁ = 100 cm³; V₂ = ?
T₁ = 0 ° C = 273{ K}; T₂ = {6}/{5} times 273 = 327.6{ K}
Now {P₁V₁}/{T₁} = {P₂V₂}/{T₂}
{760 times 100}/{273} = {1140 times V₂}/{327.6}
V₂ = {760 times 100 times 327.6}/{273 times 1140} = 80{ litres}
This problem combines changes in both pressure and temperature to find the final volume. It shows how the gas responds to being compressed while also being slightly warmed.
Teacher's Tip: Handle the fractions (3/2 and 6/5) first to simplify your data before plugging it into the big equation.
Exam Tip: Be careful with the units; the initial volume was in cm³, so the final answer should technically be in cm³ or converted properly.

Question 18:
Answer: 1. True
2. False
3. False
4. False
5. False
Correcting false statements is a great way to test your understanding of gas laws. For example, temperature must always be in Kelvin for these proportionalities to work correctly.
Teacher's Tip: Most "False" answers in gas laws come from ignoring the Kelvin scale or mixing up inverse and direct proportions.
Exam Tip: Read every word carefully; a single "inversely" instead of "directly" can make a statement false.

Question 19:
Answer: Boyle’s law - From Kinetic theory of gases,
{1}/{2} moverline{u}² = {3}/{2} kT
p = {1}/{3} (N/V)moverline{u}²
p = {2}/{3} (N/V){1}/{2} moverline{u}²
p = {2}/{3} (N/V)(3/2)kT
pV = NkT
Therefore p ∝ {1}/{V}
Which is Boyle’s law.
Charles’ law - From gas law,
PV = NkT
At constant pressure and constant amount of gas, volume is directly proportional to the temperature which is the statement of Charles’ law.
The Kinetic Theory of Gases provides the microscopic explanation for the laws we observe in the lab. It proves that the "pressure" we feel is actually just trillions of tiny atoms bouncing against a surface.
Teacher's Tip: "kT" represents thermal energy; more heat means more pressure and volume.
Exam Tip: Use the "PV = NkT" form to quickly derive both laws during a test.

Question 20:
Answer: Given;
V₁ = 320{ ml}; V₂ = 450{ ml}
T₁ = 47 ° C = 47 + 273 = 320{ K}; T₂ = ?
{V₁}/{T₁} = {V₂}/{T₂}
{320}/{320} = {450}/{T₂}
T₁ = 450{ K}
= (450 - 273) ° C
= 177 ° C
Since the volume increased, we knew the temperature must have increased as well. The math shows that the gas was heated from a warm 47 ° C to a very hot 177 ° C.
Teacher's Tip: Notice how V₁ and T₁ were both 320? This makes the calculation very easy!
Exam Tip: Don't forget to do the final subtraction (450 - 273) to give the answer in Celsius if required.

Question 21:
Answer: We trap a definite quantity of air in the closed vessel. At any point, the pressure on the air is equal to the atmospheric pressure plus the pressure due to the excess mercury column in the open end tube. By pouring mercury in the tube, we increase the pressure on the air and measure its volume under that pressure. We thus obtain a set of data for the volume of a fixed mass of air under different pressures. For a given mass of air at constant temperature, the following observations are made -
1. The volume of air decreases with increasing pressure and vice versa.
2. The proportion by which the volume decreases or increases is the same by which the pressure increases or decreases.
This experiment is the classic way to demonstrate Boyle's Law using a J-tube. It shows that air is highly compressible, meaning it can be squashed into a tiny space if enough force is applied.
Teacher's Tip: Mercury is used because it's heavy and doesn't evaporate easily, making it perfect for measuring pressure.
Exam Tip: Use the phrase "fixed mass of air" to show you understand that no air was added or lost during the experiment.

Question 22:
Answer: 1. Pressure will also be doubled.
2. Pressure will be double.
According to the gas laws, if you keep volume the same and double the absolute temperature, the pressure must also double. This happens because the gas particles are moving twice as fast and hitting the walls twice as often.
Teacher's Tip: Temperature and Pressure are like "best friends" - they usually go in the same direction.
Exam Tip: Use this logic to quickly check your math; if one doubles, the other should too.

Question 23:
Answer: 1. 273
2. absolute zero
3. absolute temperature
4. the average kinetic energy
These completions highlight the key terminology of gas physics. For example, temperature is scientifically defined as a measure of the average motion or kinetic energy of the particles in a substance.
Teacher's Tip: "Kinetic" means movement; so Kinetic Energy is the energy of things that move.
Exam Tip: "Absolute zero" is a specific term; always use both words together in fill-in-the-blank questions.

Question 24:
Answer: (a) According to Boyles’ law,
P₁ = 760{ mm Hg}; P₂ = ?
V₁ = 2{ litres}; V₂ = 4{ dm}³ = 4{ Litres}
P₁V₁ = P₂V₂
760 times 2 = P₂ times 4
P₂ = {760 times 2}/{4} = 380{ mm Hg}
(b) According to Boyles’ law,
P₁ = 760{ mm Hg}; P₂ = ?
V₁ = 2{ litres}; V₂ = {3}/{2} V₁ = 1.5 times 2 = 3.0{ L}
Now, P₁V₁ = P₂V₂
760 times 2 = P₂ times 3
P₂ = {760 times 2}/{3} = 506.66{ mm Hg}
These calculations show how pressure must adjust when volume is expanded or contracted. Note that 1 dm³ is exactly equal to 1 litre, which is a common conversion in chemistry problems.
Teacher's Tip: 1{ dm}³ = 1{ Litre}. Don't let the "dm" unit scare you!
Exam Tip: Round your final answer to two decimal places if the division is not even, like in part (b).

Question 25:
Answer: Given;
V₁ = 28{ cm}³; P₁ = 750{ mm Hg}
V₂ = ?; P₂ = 12{ mm Hg}
P₁V₁ = P₂V₂
750 times 28 = 12 times V₂
V₂= {750 times 28}/{12} = 1750{ cm}³
This problem shows an extreme drop in pressure, which allowed the gas to expand to more than 60 times its original size. It illustrates why gases are so useful for storing energy - they can be compressed into tiny tanks.
Teacher's Tip: Large pressure drop = Huge volume increase.
Exam Tip: Be sure to keep your initial and final units the same (cm³ stays cm³).

Question 26:
Answer: (a) Given;
P₁ = 100{ cm Hg}; T₁ = 273{ K}
P₂ = 10{ cm Hg}; T₂ = ?
{P₁}/{T₁} = {P₂}/{T₂}
{100}{273} = {10}{T₂}
T₂ = {273 times 10}/{100} = 27.3{ K}
(b) Given;
P₁ = 100{ cm Hg}; T₁ = 273{ K}
P₂ = ?; T₂ = 100 ° C = 273 + 100 = 373{ K}
{P₁}/{T₁} = {P₂}/{T₂}
{100}/{273} = {P₂}/{373}
P₂ = {373 times 100}/{273}
P₂ = 136.63{ cm Hg}
In part (a), the pressure dropped to 1/10th, so the temperature also dropped to 1/10th of its original Kelvin value. In part (b), heating the gas increased the pressure by about 36%.
Teacher's Tip: 273 K is the "freezing point of water" on the Kelvin scale.
Exam Tip: Remember to solve for T in part (a) and P in part (b) by looking at what information is "Given".

Question 27:
Answer: Given;
V₁ = 750{ cm}³; T₁ = -23 ° C = 273 - 23 = 250{ K}; P₁ = 800{ mm Hg}
V₂ = 720{ cm}³; T₂ = -3 ° C = 273 - 3 = 270{ K}; P₂ = ?
{P₁V₁}/{T₁} = {P₂V₂}/{T₂}
{800 times 750}/{250} = {P₂ times 720}/{270}
P₂ = {270 times 800 times 750}/{250 times 720} = 900{ mm Hg}
This is a full combined gas law calculation. Even though both temperature and volume changed, we can find the final pressure by organizing our data carefully into the formula.
Teacher's Tip: Be very careful when subtracting for negative Celsius temperatures (273 - 23 = 250).
Exam Tip: Write down all six variables (P₁, V₁, T₁, P₂, V₂, T₂) in a list before you start the math.

Question 28:
Answer: As weather balloon go higher into the atmosphere, the air becomes less dense, so air pressure drops. Because of this, the air that is already inside the balloon expands to cope with the difference in pressure. The end result is that the balloon expands making it larger.
This is a real-world application of Boyle's Law. As the outside "squeezing force" (pressure) decreases, the internal gas pushes outward more easily, stretching the balloon.
Teacher's Tip: This is why high-altitude balloons are often launched only partially inflated; they need room to grow!
Exam Tip: Use the words "less dense" and "pressure drops" to explain the environmental change correctly.

Question 29:
Answer: P₁ = 20{ atm}; T₁ = 27 ° C = 273 + 27 = 300{ K}
P₂ = 20 + 20 times {20}/{100} = 24{ atm}; T₂ = ?
{P₁}/{T₁} = {P₂}/{T₂}
{20}{300} = {24}{T₂}
T₂ = {300 times 24}/{20} = 360{ K}
= 360 - 273 = 87 ° C
This problem includes a percentage increase for pressure. A 20% increase on a starting pressure of 20 atm adds 4 more atm, bringing the total final pressure to 24 atm.
Teacher's Tip: 20% of 20 is 4. Add it to the original! (20 + 4 = 24).
Exam Tip: Be sure to add the increase to the original value rather than just using the increase itself in the formula.