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Thermal Properties of Matter
Physics · Class 11NCERT pp. 202–2212 sets × 25 questions
Key facts and formulas for Thermal Properties of Matter, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Temperature scales | (t_F − 32)/180 = t_C/100; K = t_C + 273.15 |
| Gas laws | Constant V: P ∝ T; constant P: V ∝ T (T in kelvin) |
| Thermal expansion | Δl = α l ΔT; α_A = 2α; α_V = γ = 3α; ideal gas α_V = 1/T |
| Thermal stress | F = Y A α ΔT when the rod is held rigidly |
| Heat and calorimetry | Q = m s ΔT; Q = m L; heat lost = heat gained; water equivalent = m s / s_water |
| Conduction | H = K A ΔT / L; series R = L/(K A) adds; parallel K_eff = (K₁A₁ + K₂A₂)/(A₁ + A₂) |
| Joint temperature | T₀ = (K₁T₁ + K₂T₂)/(K₁ + K₂) for equal rods in series |
| Radiation | Stefan: P = σ A T⁴; Wien: λ_m T = constant; good absorber = good emitter |
| Newton's cooling | Rate of fall = K (T_avg − T₀); excess = e₀ e^(−Kt); half-time = ln 2 / K |
| Water and states | Densest at 4 °C; melting point of ice falls with pressure (regelation); temperature stays constant during a change of state |
Three sample questions
Q1. A patient's temperature is 37 °C. On the Fahrenheit scale this is
- 69.4 °F
- 98.6 °F
- 100.4 °F
- 95.0 °F
Hint and solution
Hint: Use the conversion between the two scales: multiply the Celsius value by 9/5 first, then add 32.
- (t_F − 32)/180 = t_C/100 — The two scales are linear. Ice point is 0 and 32, steam point is 100 and 212, so the ratio of divisions is 180 to 100, that is 9/5.
- t_F = (9/5) × 37 + 32 — Substitute t_C = 37 into the conversion formula.
- (9/5) × 37 = 66.6; 66.6 + 32 = 98.6 — Multiply first, then add 32.
- Answer: 98.6 °F — The patient's temperature is 98.6 °F. Option B.
Answer: B. t_F = (9/5)(37) + 32 = 66.6 + 32 = 98.6 °F.
Q2. A constant-volume gas thermometer reads 80 kPa at 273.15 K. The pressure at 100 °C (373.15 K) is about
- 80 kPa
- 146 kPa
- 59 kPa
- 109 kPa
Hint and solution
Hint: At constant volume the pressure is proportional to the absolute (kelvin) temperature, so use the ratio of kelvin temperatures.
- Constant V: P₁/T₁ = P₂/T₂ — For a gas at fixed volume, pressure is directly proportional to kelvin temperature.
- P₂ = 80 × 373.15 / 273.15 — Substitute P₁ = 80 kPa, T₁ = 273.15 K and T₂ = 373.15 K.
- 373.15/273.15 = 1.366; 80 × 1.366 = 109.3 — The ratio is about 1.366, so the pressure is multiplied by this.
- Answer: about 109 kPa — The pressure at 100 °C is about 109 kPa. Option D.
Answer: D. At fixed volume P ∝ T. P = 80 × 373.15/273.15 = 109.3 kPa.
Q3. Statement I: P–t lines of different low-density gases at constant volume meet at the same point, −273.15 °C.
Statement II: A real gas actually reaches zero pressure at −273.15 °C.
Choose the correct option.
- I is true and II is false
- Both are true
- I is false and II is true
- Both are false
Hint and solution
Hint: Ask how the value −273.15 °C was found: by extending lines on a graph, or by actually cooling a gas that far? Real gases liquefy first.
- P–t lines extrapolate to −273.15 °C — For low-density gases at constant volume, P versus t is a straight line. Extending it back, every gas line hits zero pressure at −273.15 °C. So statement I is true.
- Real gases deviate at low T — Real gases condense to liquid or solid before reaching that temperature, and the straight-line law fails there. So statement II is false.
- I true, II false — So the answer is the option saying I is true and II is false. Option A.
Answer: A. Absolute zero comes from extrapolating the straight lines. Real gases depart from ideal behaviour at low temperature, so statement II is false.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.