Maharashtra Board Class 12 Physics Chapter 3 Kinetic Theory of Gases and Radiation Exercise Solutions

1. Choose the correct option.

(i) In an ideal gas, the molecules possess
(a) only kinetic energy
(b) both kinetic energy and potential energy
(c) only potential energy
(d) neither kinetic energy nor potential energy
Answer: (a) only kinetic energy
In simple words: In an ideal gas, we assume molecules don't pull or push on each other, so they only have energy because they are moving around, which is called kinetic energy.

📝 Teacher's Note: Remind students that the definition of an ideal gas assumes zero intermolecular forces, which mathematically eliminates potential energy from the system.

🎯 Exam Tip: In multiple-choice questions about ideal gases, remember "No forces = No potential energy".

(ii) The mean free path \( \lambda \) of molecules is given by
(a) \( \sqrt{\frac{2}{\pi nd^2}} \)
(b) \( \frac{1}{\pi nd^2} \)
(c) \( \frac{1}{\sqrt{2}\pi nd^2} \)
(d) \( \frac{1}{\sqrt{2\pi nd^1}} \)
where n is the number of molecules per unit volume and d is the diameter of the molecules.
Answer: (c) \( \frac{1}{\sqrt{2}\pi nd^2} \)
In simple words: The mean free path is the average distance a tiny gas particle travels before hitting another one; the formula shows that if there are more particles or they are bigger, they hit each other more often.

📝 Teacher's Note: Use a "crowded hallway" analogy to explain why density (\( n \)) and size (\( d \)) are in the denominator.

🎯 Exam Tip: Be careful with the \( \sqrt{2} \) factor; it distinguishes the simplified kinetic model from the rigorous Maxwellian distribution.

(iii) If pressure of an ideal gas is decreased by 10% isothermally, then its volume will
(a) decrease by 9%
(b) increase by 9%
(c) decrease by 10%
(d) increase by 11.11%
Answer: (d) increase by 11.11%
In simple words: When you lower the pressure on a gas without changing the temperature, the gas expands; a 10% drop in pressure requires a bit more than a 10% increase in space to keep things balanced.

📝 Teacher's Note: Show the calculation \( V_2 = \frac{P_1 V_1}{0.9 P_1} = 1.111 V_1 \) to help students see why the percentage isn't a simple subtraction/addition.

🎯 Exam Tip: For isothermal processes, always use Boyle's Law (\( P_1 V_1 = P_2 V_2 \)) and work with decimal factors rather than just percentages.

(iv) If a = 0.72 and r = 0.24, then the value of \( t_r \) is
(a) 0.02
(b) 0.04
(c) 0.4
(d) 0.2
Answer: (b) 0.04
In simple words: When light or heat hits a surface, it can be absorbed, reflected, or passed through; since the total must be 100% (or 1), we just subtract the known parts from 1 to find the missing piece.

📝 Teacher's Note: Emphasize the fundamental equation \( a + r + t = 1 \) for thermal radiation coefficients.

🎯 Exam Tip: This is an easy scoring question. Just ensure your subtraction is accurate: \( 1 - (0.72 + 0.24) = 0.04 \).

(v) The ratio of emissive power of a perfect blackbody at 1327°C and 527°C is
(a) 4 : 1
(b) 16 : 1
(c) 2 : 1
(d) 8 : 1
Answer: (b) 16 : 1
In simple words: A blackbody radiates energy based on its temperature raised to the power of 4. When we double the absolute temperature (from 800 K to 1600 K), the power jumps by \( 2^4 \), which is 16 times.

📝 Teacher's Note: Students often forget to convert Celsius to Kelvin. Always check if they added 273 before doing the ratio.

🎯 Exam Tip: Remember Stefan's Law: \( R \propto T^4 \). Converting to Kelvin is mandatory; \( 1327 + 273 = 1600 K \) and \( 527 + 273 = 800 K \).

2. Answer in brief.

(i) What will happen to the mean square speed of the molecules of a gas if the temperature of the gas increases?
Answer: If the temperature of a gas increases, the mean square speed of the molecules of the gas will increase in the same proportion.
[Note: \( \bar{v^2} = \frac{3nRT}{Nm} \therefore \bar{v^2} \propto T \) for a fixed mass of gas.]
In simple words: Temperature is just a measure of how fast molecules are vibrating or moving; so if the gas gets hotter, the molecules move faster on average.

📝 Teacher's Note: Use the "dancing molecules" analogy to explain that heat is kinetic energy at a microscopic level.

🎯 Exam Tip: State the proportionality clearly: \( \bar{v^2} \propto T \). Use the bar symbol to indicate "average" or "mean".

(ii) On what factors do the degrees of freedom depend?
Answer: The degrees of freedom depend upon
(i) the number of atoms forming a molecule
(ii) the structure of the molecule
(iii) the temperature of the gas.
In simple words: Degrees of freedom are just the number of ways a molecule can move (like sliding, spinning, or vibrating). It depends on how many atoms are stuck together and if it's hot enough to make them shake.

📝 Teacher's Note: Explain that vibrational degrees of freedom usually "unlock" only at very high temperatures.

🎯 Exam Tip: Mentioning "temperature" is crucial because it determines whether vibrational modes are active or dormant.

(iii) Write ideal gas equation for a mass of 7 g of nitrogen gas.
Answer: In the usual notation, \( PV = nRT \).
Here, \( n = \frac{\text{mass of the gas}}{\text{molar mass}} = \frac{7}{28} = \frac{1}{4} \)
Therefore, the corresponding ideal gas equation is \( PV = \frac{1}{4} RT \).
In simple words: Nitrogen molecules have a weight of 28. Since we only have 7 grams, we have a quarter of a "mole," so we put 1/4 into the standard gas formula.

📝 Teacher's Note: Ensure students know nitrogen is diatomic (\( N_2 \)), so its molar mass is \( 14 \times 2 = 28 \), not 14.

🎯 Exam Tip: Always show the calculation for 'n' to justify the final equation.

(iv) What is an ideal gas ? Does an ideal gas exist in practice ?.
Answer: An ideal or perfect gas is a gas which obeys the gas laws (Boyle’s law, Charles’ law and Gay-Lussac’s law) at all pressures and temperatures. An ideal gas cannot be liquefied by application of pressure or lowering the temperature. A molecule of an ideal gas is an ideal particle having only mass and velocity. Its structure and size are ignored. Also, intermolecular forces are zero except during collisions. In practice, no gas is perfectly ideal.
In simple words: An ideal gas is a "perfect" imaginary gas where particles don't stick together and take up no space. Real gases aren't perfect, but they act like this if they are hot and have lots of room.

📝 Teacher's Note: Contrast "ideal gas" with "real gas" by pointing out that real molecules do have volume and attractive forces.

🎯 Exam Tip: Mention the two main assumptions: no intermolecular forces and negligible molecular volume.

(v) Define athermanous substances and diathermanous substances.
Answer: 1. A substance which is largely opaque to thermal radiations, i.e., a substance which does not transmit heat radiations incident on it, is known as an athermanous substance.
2. A substance through which heat radiations can pass is known as a diathermanous substance.
In simple words: Athermanous things block heat rays (like wood), while diathermanous things let heat rays pass right through (like rock salt or dry air).

📝 Teacher's Note: Give examples like wood or water for athermanous and dry air or quartz for diathermanous to make it concrete.

🎯 Exam Tip: Use the word "transmit" in your definitions; it's the key technical term examiners look for.

Question 3. When a gas is heated its temperature increases. Explain this phenomenon based on kinetic theory of gases.
Answer: Molecules of a gas are in a state of continuous random motion. They possess kinetic energy. When a gas is heated, there is increase in the average kinetic energy per molecule of the gas. Hence, its temperature increases (the average kinetic energy per molecule being proportional to the absolute temperature of the gas).
In simple words: When you add heat, you are giving the gas molecules more energy to zip around faster. Since temperature is just a way to measure how fast they are zipping, the temperature goes up.

📝 Teacher's Note: Reinforce the concept that temperature is a macroscopic property representing microscopic kinetic energy.

🎯 Exam Tip: Link "Heating" \( \rightarrow \) "Increase in Kinetic Energy" \( \rightarrow \) "Increase in Temperature" in your explanation.

Question 4. Explain, on the basis of kinetic theory, how the pressure of gas changes if its volume is reduced at constant temperature.
Answer: The average kinetic energy per molecule of a gas is constant at constant temperature. When the volume of a gas is reduced at constant temperature, the number of collisions of gas molecules per unit time with the walls of the container increases. This increases the momentum transferred per unit time per unit area, i.e., the force exerted by the gas on the walls. Hence, the pressure of the gas increases.
In simple words: If you squeeze a gas into a smaller box, the molecules have less space to move, so they bang into the walls much more often. More hits on the wall mean more pressure.

📝 Teacher's Note: Use the "crowded room" analogy: if the room shrinks but people keep walking at the same speed, they will bump into the walls more frequently.

🎯 Exam Tip: Use the phrase "number of collisions per unit time" to explain the microscopic cause of pressure change.

Question 5. Mention the conditions under which a real gas obeys ideal gas equation.
Answer: A real gas obeys ideal gas equation when temperature is very high and pressure is very low. [ Note : Under these conditions, the density of a gas is very low. Hence, the molecules, on an average, are far away from each other. The intermolecular forces are then not of much consequence. ]
In simple words: Real gases act like ideal gases when they are hot and under low pressure because the molecules are zipping too fast and are too far apart to "stick" to each other.

📝 Teacher's Note: Explain that at high temp/low pressure, the deviations caused by volume and attraction become statistically negligible.

🎯 Exam Tip: "High Temperature" and "Low Pressure" are the two key phrases. Writing both is essential for full marks.

Question 6. State the law of equipartition of energy and hence calculate molar specific heat of mono-and di-atomic gases at constant volume and constant pressure.
Answer: Law of equipartition of energy: For a gas in thermal equilibrium at absolute temperature T, the average energy for a molecule, associated with each quadratic term (each degree of freedom), is \( \frac{1}{2} k_B T \), where \( k_B \) is the Boltzmann constant.
(a) Monatomic gas: Each atom has 3 degrees of freedom (translational).
Average energy per atom = \( \frac{3}{2} k_B T \). Total internal energy per mole \( E = \frac{3}{2} N_A k_B T = \frac{3}{2} RT \).
\( C_V = \frac{dE}{dT} = \frac{3}{2} R \)
\( C_P = C_V + R = \frac{3}{2} R + R = \frac{5}{2} R \)
(b) Diatomic gas (rigid): Each molecule has 5 degrees of freedom (3 trans + 2 rot).
Average energy per molecule = \( 3(\frac{1}{2} k_B T) + 2(\frac{1}{2} k_B T) = \frac{5}{2} k_B T \).
Internal energy per mole \( E = \frac{5}{2} RT \).
\( C_V = \frac{5}{2} R \) and \( C_P = \frac{7}{2} R \).
In simple words: This law says that heat energy gets shared equally among all the ways a molecule can move. Each way gets a tiny slice of energy (\( \frac{1}{2} kT \)). Since different molecules move in different ways, they soak up heat differently.

📝 Teacher's Note: Use a budget analogy: if you have $300 and 3 kids, each gets $100. Here, the energy is the budget and degrees of freedom are the kids.

🎯 Exam Tip: Always state Mayer's relation (\( C_P - C_V = R \)) when calculating \( C_P \).

Question 7. What is a perfect blackbody ? How can it be realized in practice?
Answer: A perfect blackbody is defined as a body which absorbs all the radiant energy incident on it. In practice, it can be realized using Fery’s blackbody. It consists of a hollow double-walled metallic sphere with a tiny hole. The inside is blackened with lampblack and has a conical projection opposite the hole to prevent radiation from reflecting straight back out. Any light entering the hole gets trapped by multiple internal reflections until it is almost 100% absorbed.
In simple words: A perfect blackbody is like a bottomless pit for light; nothing bounces back. Scientists build one using a hollow ball with a tiny hole and black paint inside so light gets lost bouncing around inside.

📝 Teacher's Note: Explain that the "blackness" is due to the geometry (the hole), not just the paint.

🎯 Exam Tip: Drawing the diagram with the conical projection is vital. Label the "lampblack" coating and the "aperture".

Question 8. State (i) Stefan-Boltzmann law and (ii) Wein’s displacement law.
Answer: (i) The Stefan-Boltzmann law: The rate of emission of radiant energy per unit area of a perfect blackbody is directly proportional to the fourth power of its absolute temperature. \( R = \sigma T^4 \).
(ii) Wien’s displacement law: The wavelength for which the emissive power of a blackbody is maximum (\( \lambda_{max} \)) is inversely proportional to its absolute temperature. \( \lambda_{max} T = b \), where b is Wien's constant.
In simple words: Stefan's law says that hotter things glow way brighter. Wien's law says that as things get hotter, the color of their glow shifts from red to blue/white.

📝 Teacher's Note: Use the example of a heating iron rod: it glows dull red (cool), then bright yellow, then white (very hot) to illustrate both laws.

🎯 Exam Tip: State the units for the constants: \( \sigma = 5.67 \times 10^{-8} \, J \, s^{-1} \, m^{-2} \, K^{-4} \) and \( b = 2.898 \times 10^{-3} \, m \cdot K \).

Question 9. Explain spectral distribution of blackbody radiation.
Answer: Blackbody radiation extends over a wide range of wavelengths. The spectral distribution shows how energy is spread across these wavelengths at different temperatures. 1. At any temperature, the radiation is not uniform. 2. As temperature increases, the total energy radiated (area under the curve) increases rapidly. 3. The peak of the curve shifts towards shorter wavelengths as temperature increases (Wien's Law). 4. For every wavelength, the emissive power increases with temperature.

In simple words: This graph shows that hot objects give off all kinds of light (like infrared and visible), but there's always one specific "color" they give off most. The hotter they are, the more energy they give off and the "bluer" that peak color becomes.

📝 Teacher's Note: Emphasize that these curves were the starting point for Quantum Mechanics because classical physics couldn't explain them.

🎯 Exam Tip: Draw at least two curves for different temperatures \( T_1 \) and \( T_2 \) (\( T_1 > T_2 \)) to show the shift in the peak.

Question 10. State and prove Kirchhoff’s law of heat radiation.
Answer: Law: For a body in thermal equilibrium, its emissivity (e) is equal to its absorptivity (a). i.e., \( a = e \).
Proof: Consider an ordinary body A and a perfect blackbody B inside an enclosure. At equilibrium, they reach the same temperature. For body A, heat absorbed = heat radiated. \( aQ = R \). For blackbody B, all heat is absorbed, so \( Q = R_b \). Dividing these: \( \frac{aQ}{Q} = \frac{R}{R_b} \implies a = \frac{R}{R_b} \). By definition, \( \frac{R}{R_b} = e \). Therefore, \( a = e \).

In simple words: This law says that if a material is good at soaking up heat, it must also be just as good at giving it off. A perfect sponge is also a perfect sprayer.

📝 Teacher's Note: Use the "good absorber is a good emitter" catchphrase. It's why dark clothes feel hot in the sun but also cool down faster at night.

🎯 Exam Tip: Don't forget to define \( R_b \) as the emissive power of a perfect blackbody.

Question 11. Calculate the ratio of mean square speeds of molecules of a gas at 30 K and 120 K.
Answer: Data: \( T_1 = 30 \, K \), \( T_2 = 120 \, K \)
The mean square speed, \( \bar{v^2} = \frac{3RT}{M_0} \)
\( \implies \frac{\bar{v_1^2}}{\bar{v_2^2}} = \frac{T_1}{T_2} \) for a given gas.
\( \implies \frac{\bar{v_1^2}}{\bar{v_2^2}} = \frac{30 \, K}{120 \, K} = \frac{1}{4} \)
The required ratio is 1:4.
In simple words: Since square speed is directly linked to temperature, if the temperature is 4 times higher, the square of the speed is also 4 times higher.

📝 Teacher's Note: Clarify the difference between "mean square speed" (\( \bar{v^2} \)) and "root mean square speed" (\( v_{rms} \)).

🎯 Exam Tip: The question asks for the ratio of mean square speeds, not rms speeds. Don't take the square root of 1/4!

Question 12. Two vessels A and B are filled with same gas where volume, temperature and pressure in vessel A is twice the volume, temperature and pressure in vessel B. Calculate the ratio of number of molecules of gas in vessel A to that in vessel B.
Answer: Data: \( V_A = 2V_B \), \( T_A = 2T_B \), \( P_A = 2P_B \)
Using \( PV = Nk_B T \), we have \( N = \frac{PV}{k_B T} \)
\( \implies \frac{N_A}{N_B} = (\frac{P_A}{P_B}) \times (\frac{V_A}{V_B}) \times (\frac{T_B}{T_A}) \)
\( \implies \frac{N_A}{N_B} = (2) \times (2) \times (\frac{1}{2}) = 2 \)
The ratio is 2:1.
In simple words: Even though vessel A is bigger and has more pressure (which adds molecules), it's also hotter (which pushes them out). After balancing these, A ends up with exactly twice as many molecules as B.

📝 Teacher's Note: Show students how to set up the ratio calculation by grouping similar variables. It prevents calculation errors.

🎯 Exam Tip: Be careful with the temperature ratio; since \( T \) is in the denominator, it's \( T_B / T_A \), not \( T_A / T_B \).

Question 13. A gas in a cylinder is at pressure P. If the masses of all the molecules are made one third of their original value and their speeds are doubled, then find the resultant pressure.
Answer: Data: \( m_2 = \frac{m_1}{3} \), \( v_{rms2} = 2v_{rms1} \)
Pressure \( P = \frac{1}{3} \frac{mN}{V} v_{rms}^2 \)
\( \implies \frac{P_2}{P_1} = (\frac{m_2}{m_1}) \times (\frac{v_{rms2}^2}{v_{rms1}^2}) \)
\( \implies \frac{P_2}{P_1} = (\frac{1}{3}) \times (2)^2 = \frac{4}{3} \)
Resultant pressure \( P_2 = \frac{4}{3} P \).
In simple words: Making molecules lighter lowers pressure, but making them faster increases it. Since speed is squared in the formula, the "faster" part wins out, making the pressure higher than it started.

📝 Teacher's Note: Emphasize that pressure depends on the square of velocity, so doubling speed has a massive effect (4x).

🎯 Exam Tip: Clearly list the initial and final variables (\( m_1, m_2, v_1, v_2 \)) to avoid mixing them up in the ratio.

Question 14. Show that rms velocity of an oxygen molecule is \( \sqrt{2} \) times that of a sulfur dioxide molecule at S.T.P.
Answer: Data: \( M_0(SO_2) = 64 \, g/mol \), \( M_0(O_2) = 32 \, g/mol \)
\( v_{rms} = \sqrt{\frac{3RT}{M_0}} \)
At constant T, \( v_{rms} \propto \frac{1}{\sqrt{M_0}} \)
\( \implies \frac{v_{rms}(O_2)}{v_{rms}(SO_2)} = \sqrt{\frac{M_0(SO_2)}{M_0(O_2)}} = \sqrt{\frac{64}{32}} = \sqrt{2} \)
\( \implies v_{rms}(O_2) = \sqrt{2} v_{rms}(SO_2) \).
In simple words: Lighter things move faster. Oxygen is half as heavy as Sulfur Dioxide, so it zips around \( \sqrt{2} \) (about 1.4) times faster.

📝 Teacher's Note: Use the "heavy runner vs light runner" analogy. If both have the same energy, the lighter one must be faster.

🎯 Exam Tip: Mention that both gases are at the same temperature (STP) to justify the constant T assumption.

Question 15. At what temperature will oxygen molecules have same rms speed as helium molecules at S.T.P.? (Molecular masses of oxygen and helium are 32 and 4 respectively)
Answer: Data: \( T_2 = 273 \, K \), \( M_{O2} = 32 \), \( M_{He} = 4 \)
For same rms speed, \( \frac{T_1}{M_{O2}} = \frac{T_2}{M_{He}} \)
\( \implies T_1 = \frac{M_{O2}}{M_{He}} \times T_2 = \frac{32}{4} \times 273 = 8 \times 273 = 2184 \, K \).
The temperature is 2184 K.
In simple words: Helium is very light and zips fast even when cold. To make heavy Oxygen move just as fast, you have to heat it up significantly—up to over 2000 degrees!

📝 Teacher's Note: This problem demonstrates the relationship \( T \propto M \) for a constant rms speed.

🎯 Exam Tip: Always state your final answer with the unit 'K'. Converting back to Celsius is not required unless asked.

Question 16. Compare the rms speed of hydrogen molecules at 127 ºC with rms speed of oxygen molecules at 27 ºC given that molecular masses of hydrogen and oxygen are 2 and 32 respectively.
Answer: Data: \( T_H = 400 \, K \), \( T_O = 300 \, K \), \( M_H = 2 \), \( M_O = 32 \)
\( \frac{v_{rmsH}}{v_{rmsO}} = \sqrt{(\frac{T_H}{T_O}) \times (\frac{M_O}{M_H})} \)
\( \implies \text{Ratio} = \sqrt{(\frac{400}{300}) \times (\frac{32}{2})} = \sqrt{\frac{4}{3} \times 16} = \sqrt{\frac{64}{3} } = \frac{8}{\sqrt{3}} \)
The ratio is \( 8 : \sqrt{3} \).
In simple words: Hydrogen is much lighter and it's hotter than the Oxygen, so it's zipping much faster. For every step an Oxygen molecule takes, the Hydrogen molecule has covered about 4.6 steps.

📝 Teacher's Note: Help students simplify fractions inside the square root before trying to solve it. It makes the math cleaner.

🎯 Exam Tip: Convert 127 ºC and 27 ºC to Kelvin immediately. Using Celsius in the ratio will result in a completely wrong answer.

Question 17. Find kinetic energy of 5000 cc of a gas at S.T.P. given standard pressure is \( 1.013 \times 10^5 \, N/m^2 \).
Answer: Data: \( P = 1.013 \times 10^5 \, N/m^2 \), \( V = 5000 \, cc = 5 \times 10^{-3} \, m^3 \)
Kinetic Energy \( E = \frac{3}{2} PV \)
\( \implies E = \frac{3}{2} (1.013 \times 10^5) (5 \times 10^{-3}) \)
\( \implies E = 7.5 \times 1.013 \times 10^2 = 759.75 \, J \approx 7.597 \times 10^2 \, J \).
In simple words: We find the total energy of the gas by multiplying the pressure and the volume it takes up, then adjusting by a factor of 1.5.

📝 Teacher's Note: Remind students that \( 1 \, cc = 10^{-6} \, m^3 \). This is a common conversion trap.

🎯 Exam Tip: The formula \( E = \frac{3}{2} PV \) is the standard way to find internal energy of a monatomic gas or the total kinetic energy of any gas at low temperatures.

Question 18. Calculate the average molecular kinetic energy (i) per kmol (ii) per kg (iii) per molecule of oxygen at 127 ºC, given that molecular weight of oxygen is 32, R is 8.31 J mol-1 K-1 and Avogadro’s number NA is 6.02 × 1023 molecules mol-1.
Answer: Data: \( T = 400 \, K \), \( M = 32 \, kg/kmol \), \( R = 8.31 \)

(i) Energy per kmol = \( \frac{3}{2} RT \times 1000 = \frac{3}{2} (8.31)(400)(1000) = 4.986 \times 10^6 \, J/kmol \).

(ii) Energy per kg = \( \frac{\text{Energy per kmol}}{M} = \frac{4.986 \times 10^6}{32} = 1.558 \times 10^5 \, J/kg \).

(iii) Energy per molecule = \( \frac{3}{2} k_B T = \frac{3}{2} (\frac{R}{N_A}) T = \frac{3}{2} (\frac{8.31}{6.02 \times 10^{23}}) 400 = 8.282 \times 10^{-21} \, J \).
In simple words: A huge group of molecules (kmol) has a lot of energy, but a single molecule has an incredibly tiny amount—so small it needs a negative power of 10 to write it!

📝 Teacher's Note: Distinguish between the gas constant \( R \) (for moles) and the Boltzmann constant \( k_B \) (for single molecules).

🎯 Exam Tip: Pay attention to "per kmol" vs "per mol". One kmol is 1000 moles, so multiply the standard \( \frac{3}{2} RT \) by 1000.

Question 19. Calculate the energy radiated in one minute by a blackbody of surface area 100 cm2 when it is maintained at 227ºC. (Take Stefen’s constant \( \sigma = 5.67 \times 10^{-8} \, J \, m^{-2} \, s^{-1} \, K^{-4} \))
Answer: Data: \( t = 60 \, s \), \( A = 10^{-2} \, m^2 \), \( T = 500 \, K \)
Energy \( Q = \sigma AT^4 t \)
\( \implies Q = (5.67 \times 10^{-8}) (10^{-2}) (500)^4 (60) \)
\( \implies Q = 5.67 \times 10^{-10} \times 625 \times 10^8 \times 60 \)
\( \implies Q = 2126.25 \, J \).
In simple words: We calculate how much heat glows off the surface by using its size, how long it's glowing, and its temperature raised to the power of 4.

📝 Teacher's Note: Show the power calculation \( 500^4 = (5 \times 10^2)^4 = 625 \times 10^8 \). Large powers often intimidate students.

🎯 Exam Tip: Ensure the time 't' is in seconds (1 minute = 60 seconds) and area 'A' is in \( m^2 \).

Question 20. Energy is emitted from a hole in an electric furnace at the rate of 20 W, when the temperature of the furnace is 727 ºC. What is the area of the hole? (Take Stefan’s constant \( \sigma \) to be \( 5.7 \times 10^{-8} \, J \, s^{-1} \, m^{-2} \, K^{-4} \))
Answer: Data: \( \frac{Q}{t} = 20 \, W \), \( T = 1000 \, K \), \( \sigma = 5.7 \times 10^{-8} \)
Rate of emission \( \frac{Q}{t} = \sigma AT^4 \)
\( \implies A = \frac{Q/t}{\sigma T^4} = \frac{20}{(5.7 \times 10^{-8})(1000)^4} \)
\( \implies A = \frac{20}{5.7 \times 10^{-8} \times 10^{12}} = \frac{20}{5.7 \times 10^4} = 3.509 \times 10^{-4} \, m^2 \).
The area of the hole is \( 3.509 \times 10^{-4} \, m^2 \).
In simple words: A furnace hole glows like a blackbody. If we know the power coming out and how hot it is, we can work backward to figure out how big the hole is.

📝 Teacher's Note: A hole in a furnace is the classic practical example of a blackbody.

🎯 Exam Tip: Power (Watt) is the same as \( Q/t \). Use this relationship directly in Stefan's law formula.

Question 21. The emissive power of a sphere of area \( 0.02 \, m^2 \) is \( 0.5 \, kcal \, s^{-1} \, m^{-2} \). What is the amount of heat radiated by the spherical surface in 20 second?
Answer: Data: \( R = 0.5 \, kcal \, s^{-1} \, m^{-2} \), \( A = 0.02 \, m^2 \), \( t = 20 \, s \)
Heat radiated \( Q = R \times A \times t \)
\( \implies Q = 0.5 \times 0.02 \times 20 = 0.2 \, kcal \).
The heat radiated is 0.2 kcal.
In simple words: Emissive power tells us how much heat comes off one square meter every second. We just multiply that by the actual size of our ball and the total time.

📝 Teacher's Note: This question uses "kcal" units. Students don't need to convert to Joules unless the final answer is requested in SI units.

🎯 Exam Tip: The units for emissive power (\( R \)) are often given in different forms; always check if it's per second or per minute.

Question 22. Compare the rates of emission of heat by a blackbody maintained at 727ºC and at 227ºC, if the black bodies are surrounded by an enclosure (black) at 27ºC. What would be the ratio of their rates of loss of heat ?
Answer: Data: \( T_1 = 1000 \, K \), \( T_2 = 500 \, K \), \( T_0 = 300 \, K \)
(i) Rate of emission \( \propto T^4 \). Ratio = \( \frac{1000^4}{500^4} = 2^4 = 16 \).
(ii) Rate of loss of heat \( \propto (T^4 - T_0^4) \)
Ratio = \( \frac{1000^4 - 300^4}{500^4 - 300^4} = \frac{10^4(100^4 - 3^4)}{10^4(5^4 - 3^4)} \text{ (after simplifying factors)} \)
\( \implies \text{Ratio} = \frac{10000 - 81}{625 - 81} = \frac{9919}{544} \approx 18.23 : 1 \).
In simple words: Objects give off heat based on their own temperature, but they "lose" heat based on the difference between them and the room. The hot body loses heat much more aggressively than the warm one.

📝 Teacher's Note: Clarify the difference between "rate of emission" (purely internal) and "rate of loss" (depends on the environment).

🎯 Exam Tip: When calculating "loss of heat," always use the formula \( \sigma A (T^4 - T_0^4) \).

Question 23. Earth’s mean temperature can be assumed to be 280 K. How will the curve of blackbody radiation look like for this temperature? Find out \( \lambda_{max} \). In which part of the electromagnetic spectrum, does this value lie? (Take Wien’s constant \( b = 2.897 \times 10^{-3} \, m \cdot K \))
Answer: Data: \( T = 280 \, K \), \( b = 2.897 \times 10^{-3} \, m \cdot K \)
\( \lambda_{max} = \frac{b}{T} = \frac{2.897 \times 10^{-3}}{280} = 1.035 \times 10^{-5} \, m \).
The curve will have the same general shape as a blackbody spectrum with a peak at \( 1.035 \times 10^{-5} \, m \). This value lies in the infrared region of the electromagnetic spectrum.
In simple words: The Earth is "warm" but not "hot" like the sun, so it mostly gives off invisible heat rays (infrared) instead of visible light.

📝 Teacher's Note: Use this to explain the greenhouse effect—incoming visible light from the sun vs. outgoing infrared from the earth.

🎯 Exam Tip: Memorize the wavelength ranges. Anything around \( 10^{-5} \, m \) or \( 10 \, \mu m \) is firmly in the infrared region.

Question 24. A small-blackened solid copper sphere of radius 2.5 cm is placed in an evacuated chamber. The temperature of the chamber is maintained at 100 ºC. At what rate energy must be supplied to the copper sphere to maintain its temperature at 110 ºC? (Take Stefan’s constant \( \sigma = 5.670 \times 10^{-8} \, J \, s^{-1} \, m^{-2} \, K^{-4} \), \( \pi = 3.1416 \))
Answer: Data: \( r = 0.025 \, m \), \( T_0 = 373 \, K \), \( T = 383 \, K \)
Rate of heat loss \( \frac{dQ}{dt} = \sigma A(T^4 - T_0^4) = \sigma (4\pi r^2)(T^4 - T_0^4) \)
\( \implies \text{Rate} = (5.67 \times 10^{-8}) \times 4 \times 3.1416 \times (0.025)^2 \times (383^4 - 373^4) \)
\( \implies \text{Rate} = 0.9624 \, W \).
Energy must be supplied at a rate of 0.9624 W.
In simple words: To keep something hot in a cooler room, you have to pump in exactly as much energy as it is leaking out into the air.

📝 Teacher's Note: Explain that "energy must be supplied" is equal to the "net rate of heat loss" to maintain steady state.

🎯 Exam Tip: Don't forget to calculate the area of the sphere using \( 4\pi r^2 \). Using \( \pi r^2 \) is a common mistake.

Question 25. Find the temperature of a blackbody if its spectrum has a peak at (a) \( \lambda_{max} = 700 \, nm \) (visible), (b) \( \lambda_{max} = 3 \, cm \) (microwave region) and (c) \( \lambda_{max} = 3 \, m \) (short radio waves).
Answer: Using Wien's Law: \( T = \frac{b}{\lambda_{max}} \)

(a) \( T = \frac{2.897 \times 10^{-3}}{700 \times 10^{-9}} = 4138 \, K \).

(b) \( T = \frac{2.897 \times 10^{-3}}{3 \times 10^{-2}} = 0.09657 \, K \).

(c) \( T = \frac{2.897 \times 10^{-3}}{3} = 0.0009657 \, K = 9.657 \times 10^{-4} \, K \).
In simple words: Short wavelengths (like visible light) come from very hot things, while extremely long wavelengths (like radio waves) come from things that are almost at absolute zero.

📝 Teacher's Note: This exercise shows how temperature determines the "nature" of radiation emitted by cosmic objects.

🎯 Exam Tip: Convert all lengths to meters before dividing. \( 700 \, nm = 700 \times 10^{-9} \, m \) and \( 3 \, cm = 0.03 \, m \).

Remember This (Textbook Page No. 60)

Question 1. Distribution of speeds of molecules of a gas.
Answer: Maxwell-Boltzmann distribution of molecular speeds is a relation that describes the distribution of speeds among the molecules of a gas at a given temperature. The root-mean-square speed \( v_{rms} \) gives us a general idea of molecular speeds in a gas at a given temperature. However, not all molecules have the same speed. At any instant, some molecules move slowly and some very rapidly. In classical physics, molecular speeds may be considered to cover the range from 0 to \( \infty \). The molecules constantly collide with each other and with the walls of the container and their speeds change on collisions. Also the number of molecules under consideration is very large statistically. Hence, there is an equilibrium distribution of speeds. If \( dN_v \) represents the number of molecules with speeds between \( v \) and \( v + dv \), \( dN_v \) remains fairly constant at equilibrium. We consider a gas of total \( N \) molecules. Let \( \eta_v dv \) be the probability that a molecule has its speed between \( v \) and \( v + dv \). Then, \( dN_v = N \eta_v dv \) so that the fraction, i.e., the relative number of molecules with speeds between \( v \) and \( v + dv \) is \( dN_v / N = \eta_v dv \).

In simple words: Inside a gas, molecules don't all move at the same speed. Some are slow and some are super fast. This graph shows that most molecules move at a "middle" speed, and very few are extremely slow or extremely fast.

📝 Teacher's Note: Explain that the area under the entire curve represents the total number of molecules. The peak of the curve is the "most probable speed."

🎯 Exam Tip: Remember that as temperature increases, the peak of the Maxwell-Boltzmann curve shifts to the right and becomes flatter.

Remember This (Textbook Page No. 64)

Question 1. If a hot body and a cold body are kept in vacuum, separated from each other, can they exchange heat ? If yes, which mode of transfer of heat causes change in their temperatures ? If not, give reasons.
Answer: Yes. Radiation.
In simple words: Even in empty space (vacuum) where there is no air, heat can still travel. It travels as "radiation" rays, just like how sunlight reaches the Earth through the vacuum of space.

📝 Teacher's Note: Use the sun-earth example to clarify that unlike conduction and convection, radiation does not require a material medium.

🎯 Exam Tip: Explicitly mention "Radiation" as the mode of transfer to get full marks for this conceptual question.

Remember This (Textbook Page No. 66)

Question 1. Can a perfect blackbody be realized in practice ?
Answer: For almost all practical purposes, Fery’s blackbody is very close to a perfect blackbody.
In simple words: Nothing in the real world is 100% perfect, but scientists can build a special device called Fery's blackbody that acts almost exactly like a perfect one.

📝 Teacher's Note: Explain that a "perfect" blackbody is an ideal concept, similar to the "ideal gas" concept discussed earlier.

🎯 Exam Tip: Mention "Fery's blackbody" as the closest practical approximation of an ideal blackbody.

Question 2. Are good absorbers also good emitters ?
Answer: Yes.
In simple words: This is Kirchhoff's Law. If a material is really good at soaking up heat, it is also really good at giving it back out.

📝 Teacher's Note: Remind students of the logic \( a = e \). A body that can't absorb energy efficiently can't emit it efficiently either in equilibrium.

🎯 Exam Tip: This is a frequent one-mark question. A simple "Yes" followed by "according to Kirchhoff's Law" is ideal.

Use Your Brain Power (Textbook Page No. 68)

Question 1. Why are the bottom of cooking utensils blackened and tops polished ?
Answer: The bottoms of cooking utensils are blackened to increase the rate of absorption of radiant energy and tops are polished to increase the reflection of radiation.
In simple words: We want the bottom of the pan to suck in as much heat as possible from the stove, so we paint it black. We want the top to stay shiny so it reflects heat back inside and doesn't let it escape.

📝 Teacher's Note: Connect this to the previous question: black is a good absorber, while shiny/polished surfaces are good reflectors (poor absorbers/emitters).

🎯 Exam Tip: Use the keywords "absorption" for the black bottom and "reflection" for the polished top.

Question 2. A car is left in sunlight with all its windows closed on a hot day. After some time it is observed that the inside of the car is warmer than the outside air. Why?
Answer: The air inside the car is trapped and hence is a bad conductor of heat.
In simple words: The sun's heat rays get inside through the glass, but the hot air gets trapped inside. Since air doesn't move heat through itself very well, the heat just builds up like an oven.

📝 Teacher's Note: This is a practical example of the Greenhouse Effect. Short-wavelength radiation enters, but long-wavelength radiation (heat) is trapped by the glass and the air.

🎯 Exam Tip: Focus your answer on the trapping of air and the fact that air is a poor conductor of heat.

Question 3. If surfaces of all bodies are continuously emitting radiant energy, why do they not cool down to 0 K?
Answer: Bodies absorb radiant energy from their surroundings.
In simple words: Objects are always playing a "heat exchange" game. They give out heat, but they also take in heat from the walls, the floor, and the air around them. As long as they take in as much as they give out, they don't get colder.

📝 Teacher's Note: Explain Prevost's Theory of Heat Exchange, which states that emission and absorption happen simultaneously.

🎯 Exam Tip: Mention the term "thermal equilibrium" if the body's temperature remains constant.

Can You Tell? (Textbook Page No. 71)

Question 1. \( \lambda_{max} \) the wavelength corresponding to maximum intensity for the Sun is in the blue-green region of the visible spectrum. Why does the Sun then appear yellow to us?
Answer: The colour that we perceive depends upon a number of factors such as absorption and scattering by the atmosphere (which in turn depends upon the composition of air) and the spectral response of the human eye. The colour maybe yellow/orange/red/white.
In simple words: Even though the Sun puts out most of its energy in blue-green light, our air scatters the blue light away (which makes the sky blue). What’s left over looks yellow or orange to our eyes.

📝 Teacher's Note: This is a great way to link Physics (Wien's Law) with Atmospheric Science (Rayleigh scattering).

🎯 Exam Tip: Attribute the color change to "atmospheric scattering" and "human eye sensitivity" for a complete answer.