Maharashtra Board Class 11 Chemistry Chapter 10 States of Matter Solutions

States Of Matter Class 11 Exercise Question Answers Solutions Maharashtra Board

Class 11 Chemistry Chapter 10 Exercise Solutions Maharashtra Board

Chemistry Class 11 Chapter 10 Exercise Solutions

1. Select And Write The Most Appropriate Alternatives From The Given Choices.

Question A.The unit of viscosity is
a. dynes
b. newton
c. gram
d. poise
Answer:
d. poise
In simple words: The poise is the CGS unit of dynamic viscosity, commonly used to measure how easily a fluid flows.

🎯 Exam Tip: Remember common units for physical quantities as they are frequently tested in MCQs.

Question B.Which of the following is true for 2 moles of an ideal gas ?
a. PV = nRT
b. PV = RT
c. PV = 2RT
d. PV = T
Answer:
c. PV = 2RT
In simple words: The ideal gas equation is \(PV = nRT\). If \(n = 2\) moles, then the equation becomes \(PV = 2RT\).

🎯 Exam Tip: Understand how to apply the ideal gas law \(PV = nRT\) for different numbers of moles.

Question C.Intermolecular forces in liquid are
a. greater than gases
b. less than solids
c. both a and b
d. greater than solids
Answer:
c. both a and b
In simple words: Liquids have stronger intermolecular forces than gases because their particles are closer but weaker forces than solids where particles are held in fixed positions.

🎯 Exam Tip: Compare the strength of intermolecular forces across different states of matter (solids, liquids, gases) to understand their properties.

Question D.Interactive forces are ---------- in ideal gas.
a. nil
b. small
c. large
d. same as that of real gases
Answer:
a. nil
In simple words: In an ideal gas, it is assumed that there are no attractive or repulsive forces between the gas molecules.

🎯 Exam Tip: Key assumptions of an ideal gas include negligible volume of particles and no intermolecular forces, which simplifies calculations.

Question E.At constant temperature the pressure of 22.4 dm³ volume of an ideal gas was increased from 105 kPa to 210 kPa, New volume could be-
a. 44.8 dm³
b. 11.2 dm³
c. 22.4 dm³
d. 5.6 dm³
Answer:
b. 11.2 dm³
In simple words: According to Boyle's Law, at constant temperature, pressure and volume are inversely proportional (\(P_1V_1 = P_2V_2\)). If pressure doubles from 105 kPa to 210 kPa, the volume must halve from 22.4 dm³ to 11.2 dm³.

🎯 Exam Tip: Apply Boyle's Law (\(P_1V_1 = P_2V_2\)) when temperature and amount of gas are constant to solve problems involving pressure and volume changes.

2. Answer In One Sentence.

Question A.Name the term used for mixing of different gases by random molecular motion and ferquent collision.
Answer:
The mixing of different gases by random molecular motion and frequent collision is called diffusion.
In simple words: Diffusion is the process where gas particles spread out from an area of higher concentration to an area of lower concentration, mixing completely due to their random movement and collisions.

🎯 Exam Tip: Clearly distinguish between diffusion (mixing of gases) and effusion (gas escaping through a small hole) for accurate definitions.

Question B.The pressure that each individual gas would exert if it were alone in the container, what do we call it as ?
Answer:
The pressure that each individual gas would exert if it were alone in the container is called as partial pressure.
In simple words: Partial pressure is the pressure contributed by a single gas in a mixture, as if that gas were the only one present in the container.

🎯 Exam Tip: Understand that Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of its individual components.

Question C.When a gas is heated the particles move more quickly. What is the change in volume of a heated gas if the pressure is kept constant ?
Answer:
The volume of the gas increases on heating if pressure is kept constant.
In simple words: When a gas is heated at constant pressure, its particles gain kinetic energy, move faster, and push the container walls more forcefully, leading to an increase in volume.

🎯 Exam Tip: This phenomenon is described by Charles's Law, where volume and absolute temperature are directly proportional at constant pressure.

Question D.A bubble of methane gas rises from the bottom of the North sea. What will happen to the size of the bubble as it rises to the surface ?
Answer:
According to Boyle's law, the size of the bubble of methane gas increases as it rises to the surface.
In simple words: As the bubble rises, the external water pressure decreases, causing the volume of the gas inside the bubble to expand in accordance with Boyle's Law.

🎯 Exam Tip: Boyle's Law is crucial for understanding how gas volumes change with pressure, especially in varying depth environments like water bodies.

Question E.Convert the following temperatures from degree celcius to kelvin.
a. -15° C
b. 25° C
c. -197° C
d. 273° C
Answer:
a. \(T(K) = t°C + 273.15\)
\(\therefore T(K) = -15 °C + 273.15 = 258.15 K\)
b. \(T(K) = t°C + 273.15\)
\(\therefore T(K) = 25 °C + 273.15 = 298.15 K\)
c. \(T(K) = t°C + 273.15\)
\(\therefore T(K) = -197 °C + 273.15 = 76.15 K\)
d. \(T(K) = t°C + 273.15\)
\(\therefore T(K) = 273 °C + 273.15 = 546.15 K\)
In simple words: To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature, as Kelvin is an absolute temperature scale starting from absolute zero.

🎯 Exam Tip: Always use Kelvin for gas law calculations, and remember the conversion factor is 273.15, not just 273.

Question F.Convert the following pressure values into Pascals.
a. 10 atmosphere
b. 1 kPa.
c. 107000 Nm-2
d. 1 atmosphere
Answer:
a. 10 atmosphere:
1 atm = 101325 Pa
\(\therefore 10 \text{ atm} = 1013250 \text{ Pa}\)
\(= 1.01325 \times 10^6 \text{ Pa}\)
b. 1 kPa:
1 kPa = 1000 Pa
c. 107000 N m-2:
1 N m-2 = 1 Pa
\(\therefore 107000 \text{ Nm}^{-2} = 107000 \text{ Pa}\)
\(= 1.07 \times 10^5 \text{ Pa}\)
d. 1 atmosphere:
1 atm = 101325 Pa
\(= 1.01325 \times 10^5 \text{ Pa}\)
In simple words: Pressure can be expressed in various units like atmospheres, kilopascals, or newtons per square meter (Pascals), and conversions involve specific multipliers for each unit.

🎯 Exam Tip: Familiarize yourself with common pressure units and their conversion factors, especially 1 atm = 101325 Pa, for numerical problems.

Question G.Convert:
a. Exactly 1.5 atm to pascals
b. 89 kPa to newton per square metre (N m-2)
c. 101.325 kPa to bar
d. -100 °C to Kelvin
e. 0.124 torr to standard atmosphere
Answer:
a. Exactly 1.5 atm to pascals:
1 atm = 101325 Pa
\(\therefore 1.5 \text{ atm} = 1.5 \times 101325\)
\(= 151987.5 \text{ Pa}\)
b. 89 kPa to newton per square metre (N m-2):
1 Pa = 1 N m-2 and 1 Pa = 10-3 kPa
\(\therefore 10^{-3} \text{ kPa} = 1 \text{ N m}^{-2}\)
\(\therefore 89 \text{ kPa} = \frac{1 \times 89}{10^{-3}} \text{ N m}^{-2} = 89000 \text{ N m}^{-2}\)
c. 101.325 kPa to bar:
1 bar = \(1.0 \times 10^5 \text{ Pa}\)
\(= 1.0 \times 10^2 \text{ k Pa}\)
\(\therefore 100 \text{ kPa} = 1 \text{ bar}\)
\(\therefore 101.325 \text{ kPa} = \frac{1 \times 101.325}{100}\)
\(= 1.01325 \text{ bar}\)
d. -100 °C to Kelvin:
\(T(K) = t °C + 273.15\)
\(\therefore T(K) = (- 100 °C) + 273.15 = 173.15 K\)
e. 0.124 torr to standard atmosphere:
1 atm = 760 torr
\(\therefore 1 \text{ torr} = \frac{1}{760} \text{ atm}\)
\(\therefore 0.124 \text{ torr} = 0.124 \times \frac{1}{760}\)
\(= 1.632 \times 10^{-4} \text{ atm}\)
In simple words: This question involves converting various units of pressure and temperature using standard conversion factors, which is essential for consistent calculations in chemistry.

🎯 Exam Tip: Pay close attention to unit conversions, especially for pressure (atm, kPa, Pa, torr, bar) and temperature (Celsius, Kelvin), as errors can significantly affect numerical answers.

Question H.If density of a gas is measured at constant temperature and pressure then which of the following statement is correct ?
a. Density is directly proportional to molar mass of the gas.
b. Greater the density greater is the molar mass of the gas.
c. If density, temperature and pressure is given ideal gas equation can be used to find molar mass.
d. All the above statements are correct.
Answer:
d. All the above statements are correct.
In simple words: At constant temperature and pressure, a gas's density is directly related to its molar mass, meaning denser gases have higher molar masses, and the ideal gas equation can be rearranged to calculate molar mass from density.

🎯 Exam Tip: Remember the relationship \(M = \frac{dRT}{P}\) derived from the ideal gas law, which connects molar mass (M) to density (d) at specific temperature (T) and pressure (P).

Question I.Observe the following conversions.

ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में दो H₂ अणु (दो छोटे वृत्त जुड़े हुए) और दो Cl₂ अणु (दो बड़े वृत्त जुड़े हुए) दर्शाए गए हैं, जो अभिक्रिया करके दो HCl अणु (एक छोटा और एक बड़ा वृत्त जुड़ा हुआ) बनाते हैं। दूसरे चित्र में चार H₂ अणु (चार छोटे वृत्त जुड़े हुए) और दो Cl₂ अणु (दो बड़े वृत्त जुड़े हुए) दर्शाए गए हैं, जो अभिक्रिया करके चार HCl अणु (एक छोटा और एक बड़ा वृत्त जुड़ा हुआ) बनाते हैं और दो H₂ अणु अप्रयुक्त रह जाते हैं।
Which of the above reactions is in accordance with the priciple of stoichiometry ?
Answer:
Both the reactions are in accordance with the principle of stoichiometry.
In simple words: Stoichiometry dictates the quantitative relationships between reactants and products; even if one reactant is in excess (like in the second reaction), the amount of product formed still follows the molar ratios of the limiting reactant.

🎯 Exam Tip: Stoichiometry applies even when reactants are not in perfect molar ratios; it helps identify the limiting reactant and calculate the theoretical yield of products.

In the first reaction, both the reactants are completely consumed to form product according to reaction stoichiometry.
1 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride
In the second reaction, chlorine is the limiting reagent and it is completely consumed to form hydrogen chloride. Excess hydrogen remains unreacted at the end of the reaction. This reaction also follows principle of stoichiometry.
2 mol hydrogen + 1 mol chlorine → 2 mol hydrogen chloride + 1 mol hydrogen

Question J.Hot air balloons float in air because of the low density of the air inside the balloon. Explain this with the help of an appropriate gas law.

ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में विभिन्न रंगों के कई हॉट एयर गुब्बारे खुले आसमान में उड़ते हुए दिखाए गए हैं, जो यह दर्शाते हैं कि गर्म हवा ठंडी हवा की तुलना में कम घनी होती है, जिससे वे ऊपर उठते हैं।
Answer:
The working of hot air balloon can be explained with the help of Charles' law. According to Charles' law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the air inside the balloon expands and occupies more volume. Thus, hot air inside the balloon is less dense than the surrounding cold air. This causes the hot air balloon to float in air.
In simple words: Hot air balloons operate on Charles's Law: heating the air inside increases its volume, which makes it less dense than the cooler surrounding air, causing the balloon to rise or float.

🎯 Exam Tip: Relate real-world phenomena like hot air balloons to specific gas laws (Charles's Law here) to demonstrate conceptual understanding.

3. Answer The Following Questions.

Question A.Identify the gas laws from the following diagrams.

ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र (a) में, एक पिस्टन-सिलेंडर व्यवस्था है जहाँ गैस का आयतन दबाव बढ़ने पर घटता है (1 atm से 0.5 atm तक दबाव बदलने पर आयतन बदलता है, तापमान स्थिर है)। दूसरे चित्र (b) में, एक पिस्टन-सिलेंडर व्यवस्था है जहाँ तापमान बढ़ने पर गैस का आयतन बढ़ता है (200 K से 600 K तक तापमान बदलने पर, दबाव स्थिर है)। तीसरे चित्र (c) में, समान दबाव और तापमान पर गैस के मोल की संख्या बढ़ने पर आयतन बढ़ता है (1 मोल से 1.5 मोल तक जाने पर आयतन 24.8 L से 37.2 L हो जाता है)।
Answer:
a. Boyle's law
b. Charles' law
c. Avogadro's law [Note: Assuming, T constant]
In simple words: The diagrams illustrate Boyle's Law (inverse relationship between pressure and volume at constant temperature), Charles's Law (direct relationship between volume and temperature at constant pressure), and Avogadro's Law (direct relationship between volume and moles at constant temperature and pressure).

🎯 Exam Tip: Visually identify the gas law by analyzing which variables are changing and which are held constant in the diagram, focusing on the proportional relationships.

Question B.Consider a sample of a gas in a cylinder with a movable piston.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सिलेंडर को दर्शाता है जिसमें एक चलायमान पिस्टन लगा है और नीचे गैस भरी हुई है, जो गैस के आयतन को बदलने की क्षमता को दर्शाता है।
Show digramatically the changes in the position of piston, if
a. Pressure is increased from 1.0 bar to 2.0 bar at constant temperature.
b. Temperature is decreased from 300 K to 150 K at constant pressure
c. Temperature is decreased from 400 K to 300 K and pressure is decreased from 4 bar to 3 bar.
Answer:
i.

ℹ️ चित्र व्याख्या (Diagram Explanation): पहले चित्र में, एक पिस्टन-सिलेंडर व्यवस्था है जहाँ पिस्टन ऊपर है, गैस का आयतन अधिक है, और दबाव 'constant T' पर P है। फिर दबाव को दोगुना करने पर (पिस्टन नीचे जाता है), गैस का आयतन आधा हो जाता है, जिससे पिस्टन नीचे की स्थिति में चला जाता है।
At constant T, \(P \propto \frac{1}{V}\)
Since, pressure doubles, volume will become half.
ii.

ℹ️ चित्र व्याख्या (Diagram Explanation): दूसरे चित्र में, एक पिस्टन-सिलेंडर व्यवस्था है जहाँ पिस्टन ऊपर है, गैस का आयतन अधिक है, और तापमान 'constant P' पर T है। फिर तापमान आधा करने पर (पिस्टन नीचे जाता है), गैस का आयतन भी आधा हो जाता है, जिससे पिस्टन नीचे की स्थिति में चला जाता है।
At constant P, \(V \propto T\)
Since, temperature becomes half, volume will also become half.
iii.

ℹ️ चित्र व्याख्या (Diagram Explanation): तीसरे चित्र में, एक पिस्टन-सिलेंडर व्यवस्था है जहाँ पिस्टन ऊपर है, गैस का आयतन अधिक है, और प्रारंभिक स्थिति में गैस का तापमान \(T_1\) और दबाव \(P_1\) है। फिर तापमान और दबाव में परिवर्तन के बाद (पिस्टन स्थिर रहता है), अंतिम आयतन प्रारंभिक आयतन के बराबर रहता है, जिससे पिस्टन की स्थिति में कोई बदलाव नहीं होता है।
According to combined gas law,
\(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)
Assume initial volume \(V_1 = 1 L\)
Given:
\(P_1 = 4 \text{ bar}\), \(T_1 = 400 \text{ K}\)
\(P_2 = 3 \text{ bar}\), \(T_2 = 300 \text{ K}\)
\(\therefore \frac{4 \times 1}{400} = \frac{3 \times V_2}{300}\)
\(\frac{4}{400} = \frac{3 V_2}{300}\)
\(\frac{1}{100} = \frac{V_2}{100}\)
\(\therefore V_2 = 1 L\)
Thus, the volume of the gas remains the same.
Hence, there will be no change in the position of the piston.
In simple words: These diagrams illustrate Boyle's Law (pressure and volume are inversely proportional at constant temperature), Charles's Law (volume and temperature are directly proportional at constant pressure), and the Combined Gas Law, which shows that if changes in pressure and temperature balance out, the volume can remain constant.

🎯 Exam Tip: For problems involving changes in gas properties, correctly identify the applicable gas law (Boyle's, Charles's, or Combined) and use the appropriate mathematical relationship.

 

Question C. With the help of graph answer the following -

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ स्थिर तापमान पर एक गैस के दबाव (P) और आयतन के व्युत्क्रम (1/V) के बीच संबंध दर्शाता है। जैसे-जैसे दबाव बढ़ता है, आयतन का व्युत्क्रम रैखिक रूप से बढ़ता है, जो इंगित करता है कि दबाव आयतन के व्युत्क्रमानुपाती है।
Answer:
a. \(P \propto \frac{1}{V}\)
b. The graph represents Boyle's law as it gives relation between pressure and volume at constant temperature.
c. Statement of Boyle's law: For a fixed mass (number of moles 'n') of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
OR
At constant temperature, the pressure of fixed amount (number of moles) of a gas varies inversely with its volume.
In simple words: This graph visually represents Boyle's Law, showing that when temperature is constant, gas pressure and volume are inversely related. If you increase the pressure, the volume decreases proportionally, and vice-versa.

🎯 Exam Tip: When analyzing gas law graphs, pay attention to the axes and constant variables. For Boyle's law, a P vs. 1/V graph at constant T yields a straight line through the origin, indicating inverse proportionality between P and V.

 

Question D. Write Postulates of kinetic theory of gases.

Answer: Postulates of kinetic theory of gases:
- Gases consist of tiny particles (molecules or atoms).
- On an average, gas molecules remain far apart from each other. Therefore, the actual volume of the gas molecules is negligible as compared to the volume of the container. Hence, gases are highly compressible.
- The attractive forces between the gas molecules are negligible at ordinary temperature and pressure. As a result, gas expands to occupy entire volume of the container.
- Gas molecules are in constant random motion and move in all possible directions in straight lines. They collide with each other and with the walls of the container.
- Pressure of the gas is due to the collision of gas molecules with the walls of the container.
- The collisions of the gas molecules are perfectly elastic in nature, which means that the total energy of the gaseous particle remains unchanged after collision.
- The different gas molecules move with different velocities at any instant and hence have different kinetic energies. However, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature.
In simple words: The kinetic theory of gases explains gas behavior by assuming gases are made of tiny, constantly moving particles with negligible volume and no attractive forces, whose collisions cause pressure, and whose average kinetic energy is directly related to temperature.

🎯 Exam Tip: Memorize the main postulates of the kinetic theory of gases, as they form the foundation for understanding ideal gas behavior and are frequently asked in theoretical questions.

 

Question E. Write a short note on
a. Vapour pressure.
b. Surface tension
c. Viscosity.

Answer:
a. Vapour pressure:
- Molecules of liquid have tendency to escape from its surface to form vapour above it. This called evaporation.
- When a liquid is placed in a closed container, the liquid undergoes evaporation and vapours formed undergo condensation.
- At equilibrium, the rate of evaporation and rate of condensation are equal.
- The pressure exerted by the vapour in equilibrium with the liquid is known as saturated vapour pressure or simply vapour pressure.
- Vapour pressure is measured by means of a manometer.
- The most common unit for vapour pressure is torr. 1 torr = 1 mm Hg.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक बंद बर्तन में वाष्प दाब को दर्शाता है। एक बीकर में तरल होता है, जिसके ऊपर वाष्प बनती है। वाष्प का दाब एक मैनोमीटर द्वारा मापा जाता है, जो संतुलन में तरल द्वारा लगाए गए वाष्प दाब को दर्शाता है।
b. Surface tension:
- The particles in the bulk of liquid are uniformly attracted in all directions and the net force acting on the molecules present inside the bulk is zero.
- But the molecules at the surface experience a net attractive force towards the interior of the liquid, or the forces acting on the molecules on the surface are imbalanced.
- Therefore, liquids have tendency to minimize their surface area and the surface acts as a stretched membrane.
- The force acting per unit length perpendicular to the line drawn on the surface of liquid is called surface tension.
- Unit: Surface tension is measured in SI unit, N m⁻¹ and is denoted by Greek letter 'γ'
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तरल की सतह पर और उसके अंदर अणुओं पर लगने वाले बलों को दर्शाता है। सतह पर एक अणु असंतुलित बलों का अनुभव करता है, जो उसे तरल के अंदर खींचता है, जबकि तरल के थोक में एक अणु संतुलित बलों का अनुभव करता है।
c. Viscosity:
i. Liquids (fluids) have tendency to flow.
ii. Viscosity measures the magnitude of internal friction in a liquid or fluid to flow as measured by the force per unit area resisting uniform flow.
iii. Different layers of a liquid flow with different velocity. This called laminar flow. Here, the layers of molecules in the immediate contact of the fixed surface remains stationary. The subsequent layers slip over one another. Strong intermolecular forces obstruct the layers from slipping over one another, resulting in a friction between the layers.
iv. Viscosity is defined as the force of friction between the successive layers of a flowing liquid. It is also the resistance to the flow of a liquid.
v. When a liquid flow through a tube, the central layer has the highest velocity, whereas the layer along the inner wall in the tube remains stationary. This is a result of the viscosity of a liquid. Hence, a velocity gradient exists across the cross-section of the tube.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक ट्यूब के माध्यम से तरल के लामिना के प्रवाह को दर्शाता है, जिसमें तरल की परतें अलग-अलग वेग से चलती हैं। ट्यूब के केंद्र में वेग सबसे अधिक होता है (V) और दीवारों के पास (V - dV) कम होता है, जो द्रव की चिपचिपाहट के कारण वेग ढाल को दर्शाता है।
vi. Viscosity is expressed in terms of coefficient of viscosity, 'η' (Eta). The SI unit of viscosity coefficient is N s m⁻² (newton second per square meter). In CGS system, the unit (η) is measured in poise.
1 poise = 1 g cm⁻¹ s⁻¹ = 10⁻¹ kg m⁻¹ s⁻¹
In simple words: Vapour pressure is the pressure exerted by a vapor in equilibrium with its liquid. Surface tension is the cohesive force that minimizes a liquid's surface area, acting like a stretched skin. Viscosity is a fluid's resistance to flow, caused by internal friction between its layers.

🎯 Exam Tip: When defining physical properties like vapour pressure, surface tension, and viscosity, always include their SI units and highlight the key factors influencing them (e.g., intermolecular forces for surface tension and viscosity, temperature for vapour pressure).

 

5. Solve The Following

 

Question A. A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air. As it rises in the air external pressure decreases and the volume of the gas increases till finally it bursts when external pressure is 0.3bar. What is the limit at which volume of the balloon can stay inflated ?

Answer: Given: \(P_1\) = Initial pressure = 1 bar
\(V_1\) = Initial volume = 2.27 L
\(P_2\) = Final pressure = 0.3 bar
To find: \(V_2\) = Final volume
Formula: \(P_1V_1 = P_2V_2\) (at constant n and T)
Calculation: According to Boyle's law,
\(P_1V_1 = P_2V_2\) (at constant n and T)
\( \implies V_2 = \frac{P_1 V_1}{P_2} \)
\( = \frac{1 \times 2.27}{0.3} \)
\( = 7.566667 \text{ L} \approx 7.567 \text{ L} \)
Ans: The balloon can stay inflated below the volume of 7.567 L.
In simple words: Using Boyle's law, which states that pressure and volume are inversely proportional at constant temperature, we calculated that the balloon will burst if its volume exceeds approximately 7.567 liters when the external pressure drops to 0.3 bar.

🎯 Exam Tip: For problems involving gas laws, always list the given values and the unknown quantity first. Clearly state the formula used and show all calculation steps to avoid errors and ensure full marks.

 

Question B. A syringe has a volume of 10.0 cm³ at pressure 1 atm. If you plug the end so that no gas can escape and push the plunger down, what must be the final volume to change the pressure to 3.5 atm?

Answer: Given: \(P_1\) = Initial pressure = 1 atm
\(V_1\) = Initial volume = 10.0 cm³
\(P_2\) = Final pressure = 3.5 atm
To find: \(V_2\) = Final volume
Formula: \(P_1V_1 = P_2V_2\) (at constant n and T)
Calculation: According to Boyle's law,
\(P_1V_1 = P_2V_2\) (at constant n and T)
\( \implies V_2 = \frac{P_1 V_1}{P_2} \)
\( = \frac{1 \times 10.0}{3.5} \)
\( = 2.857 \text{ L} \)
Ans: The final volume of the gas in the syringe is 2.857 L.
In simple words: By applying Boyle's Law, as the pressure inside the syringe increases from 1 atm to 3.5 atm while keeping the temperature constant, the gas volume must decrease from 10.0 cm³ to 2.857 cm³.

🎯 Exam Tip: In Boyle's Law calculations, ensure consistent units for pressure and volume. Clearly show the rearranged formula and step-by-step substitution to reach the final answer.

 

Question C. The volume of a given mass of a gas at 0°C is 2 dm³. Calculate the new volume of the gas at constant pressure when
a. The temperature is increased by 10°C.
b. The temperature is decreased by 10°C.

Answer: Given: \(T_1\) = Initial temperature = 0 °C = 0 + 273.15 = 273.15 K,
\(V_1\) = Initial volume = 2 dm³
a. \(T_2\) = Final temperature = 273.15 K + 10 = 283.15 K
b. \(T_2\) = Final temperature = 273.15 K - 10 = 263.15 K
To find: \(V_2\) = Final volume in both the cases
Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) (at constant n and P)
Calculation: According to Charles' law,
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) (at constant n and P)
i. \( \implies V_2 = \frac{V_1 T_2}{T_1} = \frac{2 \times 283.15}{273.15} = 2.073 \text{ dm}^3 \)
ii. \( \implies V_2 = \frac{V_1 T_2}{T_1} = \frac{2 \times 263.15}{273.15} = 1.927 \text{ dm}^3 \)
Ans: The new volume of a given mass of gas is:
a. 2.073 dm³
b. 1.927 dm³
In simple words: Applying Charles's Law, which states that gas volume is directly proportional to absolute temperature at constant pressure, we found that increasing the temperature by 10°C expands the gas to 2.073 dm³, while decreasing it by 10°C contracts the gas to 1.927 dm³.

🎯 Exam Tip: For Charles's Law problems, always convert temperatures from Celsius to Kelvin before calculation. Showing the two separate cases (increase and decrease in temperature) with clear steps ensures clarity and accuracy.

 

Question D. A hot air balloon has a volume of 2800 m³ at 99 °C. What is the volume if the air cools to 80 °C?

Answer: Given: \(V_1\) = Initial volume = 2800 m³, \(T_1\) = Initial temperature = 99 °C = 99 + 273.15 = 372.15 K,
\(T_2\) = Final temperature = 80 °C = 80 + 273.15 K = 353.15 K
To find: \(V_2\) = Final volume
Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) (at constant n and P)
Calculation: According to Charles' law,
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) (at constant n and P)
\( \implies V_2 = \frac{V_1 T_2}{T_1} \)
\( = \frac{2800 \times 353.15}{372.15} \)
\( = 2657 \text{m}^3 \)
Ans: The volume of the balloon when the air cools to 80 °C is 2657 m³.
In simple words: Using Charles's Law, as the air in the balloon cools from 99°C to 80°C, its volume will decrease from 2800 m³ to 2657 m³, assuming constant pressure and amount of gas.

🎯 Exam Tip: Remember to always convert Celsius temperatures to Kelvin for gas law calculations to use the absolute temperature scale, which is fundamental to Charles's Law.

 

Question E. At 0 °C, a gas occupies 22.4 liters. How much hot must be the gas in celsius and in kelvin to reach volume of 25.0 liters?

Answer: Given: \(V_1\) = Initial volume of the gas = 22.4 L,
\(T_1\) = Initial temperature = 0 + 273.15 = 273.15 K,
\(V_2\) = Final volume = 25.0 L
To find: \(T_2\) = Final temperature in Celsius and in Kelvin
Formula: \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) (at constant n and P)
Calculation: According to Charles' law,
\( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) (at constant n and P)
\( \implies T_2 = \frac{V_2 \times T_1}{V_1} \)
\( = \frac{25.0 \times 273.15}{22.4} \)
\( = 304.85 \text{ K} \approx 304.9 \text{ K} \)
Calculation using log table:
\( \frac{25.0 \times 273.15}{22.4} \)
\( = \text{Antilog}_{10} [\text{log}_{10} (25.0) + \text{log}_{10} (273.15) - \text{log}_{10} (22.4)] \)
\( = \text{Antilog}_{10} [1.3979 + 2.4365 - 1.3502] \)
\( = \text{Antilog}_{10} [2.4842] = 304.9 \)
Converting to Celsius scale:
\(T_2 = 304.85 \text{ K} - 273.15 \text{ °C} = 31.7 \text{ °C}\)
Ans: The temperature of the gas must be 31.7 °C or 304.9 K.
In simple words: To increase the gas volume from 22.4 L to 25.0 L at constant pressure, its temperature must be raised from 0°C (273.15 K) to 31.7°C (304.9 K), as per Charles's Law.

🎯 Exam Tip: When a question asks for a temperature in both Celsius and Kelvin, ensure you perform the calculation using Kelvin and then convert the final Kelvin temperature back to Celsius for the complete answer.

 

Question F. A 20 L container holds 0.650 mol of He gas at 37 °C at a pressure of 628.3 bar. What will be new pressure inside the container if the volume is reduced to 12 L. The temperature is increased to 177 °C and 1.25 mol of additional He gas was added to it?

Answer: Given: \(V_1\) = Initial volume = 20 L, \(n_1\) = Initial number of moles = 0.650 mol
\(P_1\) = Initial pressure = 628.3 bar
\(T_1\) = Initial temperature = 37 °C = 37 + 273.15 K = 310.15 K
\(n_2\) = Final number of moles = 0.650 + 1.25 = 1.90 mol, \(V_2\) = Final volume = 12 L
\(T_2\) = Final temperature = 177 °C = 177 + 273.15 K = 450.15 K, R = 0.0821 L atm K⁻¹ mol⁻¹
To find: \(P_2\) = Final pressure
Formula: PV = nRT
Calculation: According to ideal gas equation,
\(P_2V_2 = n_2RT_2\).
\( \implies P_2 = \frac{n_2 RT_2}{V_2} \)
\( = \frac{1.90 \times 0.0821 \times 450.15}{12} \)
\( = 5.852 \text{atm} \)
Ans: The final pressure of the gas is 5.852 atm.
In simple words: By using the Ideal Gas Law (PV=nRT) and accounting for changes in volume, temperature, and moles of gas, the final pressure inside the container is calculated to be 5.852 atm.

🎯 Exam Tip: For problems involving changes in multiple variables (P, V, n, T), the Ideal Gas Law (PV=nRT) is the most suitable formula. Ensure all units are consistent with the gas constant R being used, and convert temperatures to Kelvin.

 

Question G. Nitrogen gas is filled in a container of volume 2.32 L at 32 °C and 4.7 atm pressure. Calculate the number of moles of the gas.

Answer: Given: V = 2.32 L, P = 4.7 atm, T = 32 °C = 32 + 273.15 K = 305.15 K
R = 0.0821 L atm K⁻¹ mol⁻¹
To find: n = number of moles of gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
\( \implies n = \frac{PV}{RT} \)
\( = \frac{4.7 \times 2.32}{0.0821 \times 305.15} \)
\( = 0.435 \text{moles} \)
Ans: Number of moles of N2 gas in the given volume is 0.435 moles.
In simple words: Using the Ideal Gas Law (PV=nRT), we can calculate the number of moles of nitrogen gas to be 0.435, given its volume, pressure, and temperature.

🎯 Exam Tip: When calculating moles using the ideal gas law, ensure temperature is in Kelvin and pressure and volume units match the units of the gas constant R provided or selected.

 

Question H. At 25 °C and 760 mm of Hg pressure a gas occupies 600 mL volume. What will be its pressure at the height where temperature is 10 °C and volume of the gas 640 mL?

Answer: Given: \(P_1\) = Initial pressure = 760 mm Hg
\(T_1\) = Initial temperature = 25 °C = 25 + 273.15 K = 298.15 K
\(V_1\) = Initial volume = 600 mL
\(T_2\) = Final temperature = 10 °C = 10 + 273.15 K = 283.15 K
\(V_2\) = Final volume = 640 mL
To find: \(P_2\) = Final pressure
Formula: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
Calculation: According to combined gas law.
\( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \)
\( \implies P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \)
\( = \frac{760 \times 600 \times 283.15}{298.15 \times 640} \)
\( = 676.654 \text{ mm Hg} \)
Ans: The final pressure of a gas is 676.654 mm Hg.
In simple words: Using the combined gas law, which relates pressure, volume, and temperature, we find that the gas pressure decreases to 676.654 mm Hg when its temperature drops from 25°C to 10°C and its volume expands from 600 mL to 640 mL.

🎯 Exam Tip: When applying the combined gas law, ensure all temperatures are converted to Kelvin. Maintain consistency in units for pressure and volume throughout the calculation to get an accurate result.

 

Question I. A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of the gases in the cylinder is 25 bar. What is the partial pressure of dioxygen and neon in the mixture?

Answer: Given: \(m_{O_2}\) = 70.6 g, \(m_{Ne}\) = 167.5 g,
\(P_{\text{Total}}\) = 25 bar
To find: Partial pressure of each gas
Formula: \(P_1 = X_1 \times P_{\text{Total}}\)
Calculation: Determine number of moles (n) of each gas using formula: \( n = \frac{m}{M} \)
\(n_{O_2} = \frac{70.6 \text{g}}{32 \text{g mol}^{-1}} = 2.206 \text{ mol} \)
\(n_{Ne} = \frac{167.5 \text{g}}{20 \text{g mol}^{-1}} = 8.375 \text{ mol} \)
Determine the mole fraction of each gas using the formula:
\(X = \frac{n}{n_{\text{Total}}} \)
\(X_{O_2} = \frac{n_{O_2}}{n_{O_2} + n_{Ne}} = \frac{2.206 \text{ mol}}{(2.206 + 8.375) \text{ mol}} = \frac{2.206 \text{ mol}}{10.581 \text{ mol}} = 0.208 \)
\(X_{Ne} = \frac{n_{Ne}}{n_{\text{Total}}} = \frac{8.375 \text{ mol}}{10.581 \text{ mol}} = 0.792 \)
Calculate the partial pressure of each gas:
\(P_{O_2} = X_{O_2} \times P_{\text{Total}} = 0.208 \times 25 \text{ bar} = 5.2 \text{ bar} \)
\(P_{Ne} = X_{Ne} \times P_{\text{Total}} = 0.792 \times 25 \text{ bar} = 19.8 \text{ bar} \)
Ans: The partial pressure of dioxygen and neon are 5.2 bar and 19.8 bar respectively.
In simple words: To find the partial pressure of each gas in a mixture, first calculate the moles of each gas, then their mole fractions, and finally multiply each mole fraction by the total pressure of the mixture.

🎯 Exam Tip: Dalton's Law of Partial Pressures requires careful calculation of mole fractions. Ensure all molecular weights are correct for determining moles, and remember that partial pressure is the product of mole fraction and total pressure.

 

Question J. Calculate the pressure in atm of 1.0 mole of helium in a 2.0 dm³ container at 20.0 °C.

Answer: Given: n = number of moles = 1.0 mol, V = volume = 2.0 dm³
T = Temperature = 20.0 °C = 20.0 + 273.15 K = 293.15 K
R = 0.0821 L atm K⁻¹ mol⁻¹
To find: Pressure (P)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
\( \implies P = \frac{nRT}{V} \)
\( = \frac{1.0 \times 0.0821 \times 293.15}{2.0} \)
\( = 12.03 \text{ atm} \)
Calculation using log table:
\( \frac{0.0821 \times 293.15}{2.0} \)
\( = \text{Antilog}_{10} [\text{log}_{10} (0.0821) + \text{log}_{10} (293.15) - \text{log}_{10} (2.0)] \)
\( = \text{Antilog}_{10} [ \overline{2}.9143 + 2.4672 - 0.3010] \)
\( = \text{Antilog}_{10} [1.0805] = 12.03 \)
Ans: The pressure of the given helium gas is 12.03 atm.
In simple words: Using the Ideal Gas Law (PV=nRT), the pressure of 1.0 mole of helium gas in a 2.0 dm³ container at 20.0°C is calculated to be 12.03 atmospheres.

🎯 Exam Tip: Ensure that all units, especially temperature (Kelvin), are correctly applied in the Ideal Gas Law calculation. Double-check the value of the gas constant (R) and its corresponding units to match the given problem's units.

 

Question K. Calculate the volume of 1 mole of a gas at exactly 20 °C at a pressure of 101.35 kPa.

Answer: Given: n = number of moles = 1 mol, P = pressure = 101.35 kPa = 1.00025 atm ≈ 1 atm
T = Temperature = 20 °C = 20 + 273.15 K = 293.15 K
R = 0.0821 dm³ atm K⁻¹ mol⁻¹
To find: Volume (V)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
\( \implies V = \frac{nRT}{P} \)
\( = \frac{1 \times 0.0821 \times 293.15}{1} \)
\( = 24.07 \text{ dm}^3 \)
Calculation using log table:
\( \frac{0.0821}{293.15} \)
\( = \text{Antilog}_{10} [\text{log}_{10} (0.0821) - \text{log}_{10} (293.15)] \)
\( = \text{Antilog}_{10} [ \overline{2}.9143 - 2.4672] \)
\( = \text{Antilog}_{10} [1.3815] = 24.07 \)
Ans: The volume of the given gas is 24.07 dm³.
In simple words: Using the Ideal Gas Law, 1 mole of gas at 20°C and 101.35 kPa pressure will occupy a volume of approximately 24.07 dm³.

🎯 Exam Tip: When calculating volume using the ideal gas law, ensure that the pressure unit (kPa vs. atm) is consistent with the R value selected. Always convert Celsius to Kelvin for temperature.

 

Question L. Calculate the number of molecules of methane in 0.50 m³ of the gas at a pressure of 2.0 × 10² kPa and a temperature of exactly 300 K.

Answer: V = 0.5 m³, P = 2.0 × 10² kPa = 2.0 × 10⁵ Pa
T = 300 K, R = 8.314 J K⁻¹ mol⁻¹
To find: Number of molecules of methane gas
Formula: PV = nRT
Calculation: According to ideal gas equation,
\( \implies n = \frac{PV}{RT} \)
\( = \frac{2.0 \times 10^5 \times 0.5}{8.314 \times 300} \)
\( = 40 \text{ mol} \)
Number of molecules = n × NA = 40 × 6.022 × 10²³ = 2.4088 × 10²³ ≈ 2.409 × 10²⁵
Ans: The number of molecules of methane gas present is 2.409 × 10²⁵ molecules.
In simple words: First, use the Ideal Gas Law to find the moles of methane gas present. Then, multiply the number of moles by Avogadro's number to determine the total number of methane molecules.

🎯 Exam Tip: For problems involving the number of molecules, remember to use both the Ideal Gas Law to find moles and Avogadro's number to convert moles to molecules. Pay close attention to unit conversions for pressure and volume.

 

11th Chemistry Digest Chapter 10 States Of Matter Intext Questions And Answers

 

Question 1. Consider three compounds: H₂S, H₂Se and H₂O. Identify which has the highest boiling point. Justify.

Answer: Among the three compounds H₂O, H₂S and H₂Se, the first one, H₂O has the smallest molecular mass. But it has the highest B.P. of 100 °C. B.P. of H₂S is -60 °C and of H₂Se is -41.25 °C. The extraordinary high B.P. of H₂O is due to very strong hydrogen bonding even though it has the lowest molecular mass.
In simple words: Water (H₂O) has the highest boiling point among H₂O, H₂S, and H₂Se, despite its smaller molecular mass, because it forms strong hydrogen bonds, requiring more energy to break.

🎯 Exam Tip: When comparing boiling points of hydrides, always consider the presence and strength of intermolecular forces, especially hydrogen bonding, which significantly elevates boiling points beyond what molecular mass alone would predict.

 

Question i. What are the various components present in the atmosphere?

Answer: Various components present in the atmosphere are as follows:
a. Nitrogen (78%)
b. Oxygen (21%)
c. Carbon dioxide and other gases (0.03%)
d. Inert gases (mainly argon) (0.97%)
e. Traces of water vapour
In simple words: The atmosphere is mainly composed of nitrogen (78%), oxygen (21%), and smaller amounts of carbon dioxide, inert gases like argon, and water vapor.

🎯 Exam Tip: Knowing the approximate percentages of major atmospheric gases like Nitrogen and Oxygen is fundamental. Also, remember other significant minor components like Carbon dioxide and Argon.

 

Question ii. Name five elements and five compounds those exist as gases at room temperature.

Answer: Five elements and five compounds that exist as gases at room temperature are as follows:
Elements:
(a) Nitrogen
(b) Oxygen
(c) Hydrogen
(d) Chlorine
(e) Argon
Compounds:
(a) Carbon dioxide
(b) Carbon monoxide
(c) Nitrogen dioxide
(d) Sulphur dioxide
(e) Methane
In simple words: Common gaseous elements at room temperature include Nitrogen, Oxygen, Hydrogen, Chlorine, and Argon; common gaseous compounds include Carbon dioxide, Carbon monoxide, Nitrogen dioxide, Sulphur dioxide, and Methane.

🎯 Exam Tip: Be prepared to list examples of elements and compounds in different states of matter. For gaseous substances, focus on common examples like atmospheric components and simple molecular gases.

 

Question 1. What is air?

Answer:
- Air is a mixture of various gases.
- One cannot see air but can feel the cool breeze.
- The composition of air by volume is around 78 percent N₂, 21 percent O₂ and 1 percent other gases including CO₂.
In simple words: Air is an invisible mixture of gases, primarily nitrogen (78%) and oxygen (21%), along with small amounts of other gases like carbon dioxide, that we feel as a breeze.

🎯 Exam Tip: When defining air, always mention that it's a *mixture* of gases, not a single compound, and include the approximate percentages of its main components (Nitrogen and Oxygen).

 

Question 1. Find the unit in which car-tyre pressure is measured.

Answer: Car-tyre pressure is measured in the units of pounds per square inch (psi) or Newton per metre square (N m⁻²).
In simple words: Car-tyre pressure is typically measured in pounds per square inch (psi) or sometimes in Newtons per square meter (N m⁻²).

🎯 Exam Tip: Understand common pressure units for practical applications. While SI unit (Pascal or N m⁻²) is crucial for science, be aware of everyday units like psi (pounds per square inch) for car tires.

 

Question 1. How does a bicycle pump work?

Answer: A bicycle pump works on Boyle's law. Pushing a bicycle pump squashes the same number of particles into a smaller volume. This squashing means particles hit the walls of the pump more often, increasing the pressure. The increased pressure of a gas can be felt on palm by pushing in the piston of a bicycle pump.
In simple words: A bicycle pump operates based on Boyle's law; when you push the plunger, the air's volume decreases, causing its pressure to increase, forcing it into the tire.

🎯 Exam Tip: Relate everyday phenomena like a bicycle pump to fundamental gas laws. Explaining the mechanism in terms of Boyle's law (inverse relationship between pressure and volume) demonstrates a clear understanding.

 

Question 1.
i. Watch Boyle's law experiment.
ii. Find applications of Boyle's law.
iii. Try to study how Boyle's law helps in 'scuba-diving' i.e., importance of Boyle's law in scuba diving an exhilarating sport.

Answer:
i. Students can refer to 'Boyle's law experiment' on YouTube channel of 'Socratica'.
ii. a. Syringes: When the plunger of a syringe is pulled back out, it causes the volume of the gas inside it to increase due to the reduction of pressure. This creates a vacuum in the syringe, which is constantly trying to adjust the pressure back to normal. However, since the only substance available, such as the blood or medication, is on the other side of the needle, this liquid is sucked into the vacuum, increasing the pressure and decreasing the volume of the gas. When we push the plunger back down, the pressure again increases, lowering the volume inside the syringe, and pushing the fluid out.
b. Respiration: Boyle's law is essential for the human breathing process. When person breathes in, his/her lung volume increases and the pressure within decreases. Since air always moves from areas of high pressure to areas of low pressure, air is drawn into the lungs. The opposite happens when person exhales. Since the lung volume decreases, the pressure within increases, forcing the air out of the lungs
c. Storage of Gases: Many industries store gases under high pressure. This allows the gas to be stored at a low volume, saving plenty of storage space.
iii. Importance of Boyle's law in scuba diving:
a. Boyle's law affects scuba diving in many ways.
b. It explains the role of pressure in the aquatic environment.
c. As divers descend, the water pressure surrounding them increases, causing air in their body and equipment to have a smaller volume. As the divers ascend, water pressure decreases, causing their body and equipment to expand to acquire a greater volume.
d. Furthermore, it is crucial that scuba divers never attempt to hold their breath when immersed in water.
e. According to Boyle's law, if divers attempt this when they ascend to a body of water of less pressure, then the air that is trapped in their lungs will over-expand and rupture. This is known as Pulmonary Barotrauma. Thus, it is important for scuba divers to exhale as they ascend because the external pressure increases.
f. Also, if a diver returns to the surface too quickly, dissolved gases in the blood expand and form bubbles, which can get stuck in capillaries and organs (causing the 'bends').
In simple words: Boyle's law is applied in syringes for drawing/expelling fluids, in human respiration where lung volume changes alter air pressure, and crucially in scuba diving where pressure changes with depth affect gas volume in the body, necessitating controlled breathing to avoid injury.

🎯 Exam Tip: When discussing applications of Boyle's law, provide clear and concise explanations for each example. For complex scenarios like scuba diving, emphasize the safety implications and how Boyle's law governs them.

Just Think. (Textbook Page No. 145)

Question i. Why does bicycle tyre burst during summer?
Answer:

• According to Charles' law, at constant pressure, the volume of a fixed amount of a gas varies directly with the temperature. This means that as the temperature increases, the volume also increases.
• During summer, the temperature of the surrounding air is high. Due to the high temperature, the air inside the tyre gets heated. This will increase the volume of the tyres and it will burst.

In simple words: In summer, higher temperatures cause the air inside the bicycle tyre to expand, and since the tyre's volume is fixed, the increased pressure eventually leads to it bursting.

🎯 Exam Tip: Understanding the direct relationship between temperature and volume of a gas at constant pressure (Charles' Law) is key to explaining such real-world phenomena.

Question ii. Why do the hot air balloons fly high?
Answer:

• According to Charles' law, at constant pressure, gases expand on heating and become less dense. Thus, hot air is less dense than cold air.
• In a hot air balloon, the air inside it is heated by a burner. Upon heating, the air inside the balloon expands and becomes lighter (less dense) than the cooler air on the outside. This causes the hot air balloon to fly high in air.

In simple words: Hot air balloons fly because heating the air inside them makes it less dense than the cooler surrounding air, causing the balloon to float upwards due to buoyancy.

🎯 Exam Tip: Relate the concepts of Charles' law, density, and buoyancy to provide a complete explanation of how hot air balloons function for full marks.

Use Your Brainpower. (Textbook Page No. 146)

Question 1. Why does the pressure in the automobile tyres change during hot summer or winter season?
Answer:

• According to Gay-Lussac's law, at constant volume, pressure of a fixed amount of a gas is directly proportional to its absolute temperature.
• During hot summer, the temperature of automobile tyre increases faster. Consequently, the air inside the tyre gets heated and the gas molecules starts moving faster.
• As the volume of the tyre remains constant, the pressure inside it increases.
• During winter, the temperature of automobile tyre decreases. Consequently, the air inside the tyre gets cooled and the gas molecules starts moving much slower and the pressure inside the tyre decreases.

In simple words: Tyre pressure changes with seasons because gas pressure is directly proportional to temperature at constant volume (Gay-Lussac's Law); higher temperatures in summer increase pressure, while lower temperatures in winter decrease it.

🎯 Exam Tip: Clearly state Gay-Lussac's Law and explain the effect of temperature on molecular kinetic energy and collisions with the tyre walls for a comprehensive answer.

Just Think. (Textbook Page No. 149)

Question 1. Do all pure gases and mixtures of gases obey the gas laws?
Answer: Yes, the gas laws are also applicable to the mixtures of gases. The measurable properties of mixture of the gases such as pressure, temperature, volume and amount of gaseous mixture are all related by an ideal gas law.In simple words: Yes, gas laws apply to both pure gases and gas mixtures, as the collective properties of a gas mixture still follow the ideal gas equation.

🎯 Exam Tip: Remember that ideal gas laws are approximations, and real gases (pure or mixtures) may deviate at extreme conditions (high pressure, low temperature), but for general purposes, they are applicable.

Just Think. (Textbook Page No. 150)

Question 1. Where is Dalton's law applicable?
Answer: Air is gaseous mixture of different gases. Dalton's law is useful for the study of various phenomena in air, for example, air pollution.In simple words: Dalton's law is applicable to gas mixtures, like air, to understand how individual gas components contribute to the total pressure, which is useful in fields such as studying air pollution.

🎯 Exam Tip: When discussing Dalton's law, emphasize its relevance to non-reacting gas mixtures and its utility in environmental studies.

Just Think. (Textbook Page No. 155)

Question 1. What makes the oil rise through the wick in an oil lamp?
Answer: In an oil lamp, oil rises through the wick due to the capillary action. Such a capillary rise of oil is due to surface tension of oil. The wick acts as a capillary tube. When the wick is placed in oil, the attractive forces between the oil and the inner wall of capillary (wick) pull the oil up through the wick.In simple words: Oil rises in an oil lamp wick due to capillary action, which is driven by the oil's surface tension and the attractive forces between the oil and the narrow channels within the wick.

🎯 Exam Tip: Focus on explaining capillary action and surface tension as the key scientific principles behind the phenomenon for a complete answer.

11th Std Chemistry Questions And Answers