This MCQ module is based on: Thermodynamic Processes
TOPIC 16 OF 33
Thermodynamic Processes
🎓 Class 11
Physics
CBSE
Theory
Ch 11 – Thermodynamics
⏱ ~14 min
🌐 Language: [gtranslate]
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Thermodynamic Processes
11.9 Quasi-Static Processes
To apply the relation \(W = \int P\,dV\) we need pressure to be well defined throughout the process — a single value at every instant. That requires the process to be infinitely slow, so the gas is essentially in equilibrium at every instant. Such an idealised slow process is called a quasi-static process. Quasi-static processes can be plotted on a P–V diagram as smooth curves.
Real processes (a sudden expansion, an explosion) are not quasi-static, but for laboratory analysis we often treat them as such, knowing the predictions are good when the process is much slower than the molecular relaxation time.
11.10 The Five Standard Processes
Four classic processes plus a free expansion span almost every textbook problem.
11.10.1 Isobaric Process (P = constant)
Pressure is held constant. Examples: water boiling in an open pot at 1 atm; gas in a cylinder with a free piston pushed only by atmosphere.
\(W = P\,\Delta V = \mu R\,\Delta T\) (using PV = µRT).
\(\Delta U = \mu C_v\,\Delta T\); \(Q = \mu C_p\,\Delta T\). Verify \(Q = \Delta U + W\) ✓.
\(\Delta U = \mu C_v\,\Delta T\); \(Q = \mu C_p\,\Delta T\). Verify \(Q = \Delta U + W\) ✓.
11.10.2 Isochoric / Isovolumetric Process (V = constant)
Volume is held constant; e.g. gas inside a sealed rigid container being heated. Since dV = 0:
\(W = 0\); \(Q = \Delta U = \mu C_v\,\Delta T\). All heat goes into raising internal energy. P–V diagram is a vertical line.
11.10.3 Isothermal Process (T = constant)
Temperature is held constant — for example, by keeping the system in contact with a large heat reservoir. For an ideal gas \(U = U(T)\), so:
\[\Delta U = 0 \;\Longrightarrow\; Q = W\]
Using PV = µRT and integrating:
\[W = \int_{V_1}^{V_2}\frac{\mu R T}{V}dV = \mu R T \ln\frac{V_2}{V_1} = \mu R T \ln\frac{P_1}{P_2}\]
So an isothermal expansion (V₂ > V₁) absorbs exactly the same amount of heat as the work it does. The P–V curve is a hyperbola \(P = \text{const}/V\).
11.10.4 Adiabatic Process (Q = 0)
No heat exchanged with surroundings. Either the process is so fast that there is no time for heat flow, or the system is perfectly insulated. From the First Law, \(\Delta U = -W\). Combining \(dU = \mu C_v dT\) and \(\delta W = P dV = (\mu R T/V)dV\) gives:
For an ideal gas in a quasi-static adiabatic process:
\[\boxed{\,PV^\gamma = \text{const}, \quad TV^{\gamma-1} = \text{const}, \quad T^{\gamma}P^{1-\gamma} = \text{const}\,}\]
where γ = C_p/C_v. Work done by the gas:
\[W_{ad} = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{\mu R(T_1 - T_2)}{\gamma - 1}\]
An adiabatic curve on a P–V diagram (\(PV^\gamma = \text{const}\)) is steeper than the corresponding isotherm \(PV = \text{const}\). On adiabatic expansion, T falls; on adiabatic compression, T rises (the bicycle-pump effect, the diesel-engine principle).
11.10.5 Free Expansion
A gas expands into a vacuum through a partition that is suddenly removed. There is no work done (no piston pushed back, P_ext = 0) and no heat exchange (rigid insulated walls):
Q = 0, W = 0, ⇒ ΔU = 0 (ideal gas: T constant)
This is not a quasi-static process — pressure is undefined during the rush — so it cannot be plotted as a smooth curve on the P–V plane. Joule's free-expansion experiment was the historical proof that internal energy of an ideal gas depends only on T.
11.11 Summary of the Five Processes
| Process | Constraint | W (by gas) | Q | ΔU |
|---|---|---|---|---|
| Isobaric | P const | P ΔV = µR ΔT | µC_p ΔT | µC_v ΔT |
| Isochoric | V const | 0 | µC_v ΔT | µC_v ΔT |
| Isothermal | T const | µRT ln(V₂/V₁) | = W | 0 |
| Adiabatic | Q = 0 | (P₁V₁ − P₂V₂)/(γ−1) | 0 | −W |
| Cyclic | State returns | = area enclosed | = W | 0 |
Worked Examples
Example 11.8: Work in isothermal expansion (NCERT-style)
2.0 mol of an ideal gas at 27 °C expands isothermally and reversibly from a volume of 5 L to 25 L. Compute Q, W and ΔU. R = 8.314 J/mol·K.
\(W = \mu R T \ln(V_2/V_1) = 2.0\times 8.314\times 300\times \ln 5\)
\(\ln 5 = 1.609\). \(W = 2.0\times 8.314\times 300\times 1.609 = 8023\) J ≈ 8.02 kJ (work done by gas).
ΔU = 0 (isothermal, ideal gas). Q = W = 8.02 kJ.
\(\ln 5 = 1.609\). \(W = 2.0\times 8.314\times 300\times 1.609 = 8023\) J ≈ 8.02 kJ (work done by gas).
ΔU = 0 (isothermal, ideal gas). Q = W = 8.02 kJ.
Example 11.9: Adiabatic compression of air
Air (γ = 1.4) at 27 °C is compressed adiabatically and quasi-statically to 1/8 of its initial volume. Find the final temperature.
\(TV^{\gamma-1} = \text{const} \Rightarrow T_2 = T_1 (V_1/V_2)^{\gamma-1}\).
\(T_2 = 300 \times 8^{0.4} = 300 \times 2.297 = \boxed{689\text{ K}}\) ≈ 416 °C.
This is the principle of a diesel engine: air alone is compressed to ~1/15 its initial volume, raising T to ~700 °C — hot enough to ignite the diesel droplets injected at the top of the stroke.
\(T_2 = 300 \times 8^{0.4} = 300 \times 2.297 = \boxed{689\text{ K}}\) ≈ 416 °C.
This is the principle of a diesel engine: air alone is compressed to ~1/15 its initial volume, raising T to ~700 °C — hot enough to ignite the diesel droplets injected at the top of the stroke.
Example 11.10: Cyclic process with two paths (NCERT)
An ideal gas at 30 atm and 27 °C is allowed to expand isothermally to a final pressure of 1 atm. (a) What is the work done by the gas? (b) Compare to the same process done adiabatically (γ = 1.4).
(a) Isothermal: V₂/V₁ = P₁/P₂ = 30. \(W = \mu RT \ln 30\). For 1 mole: \(W = 1\times 8.314\times 300\times 3.4 = 8480\) J.
(b) Adiabatic to same P₁/P₂ = 30: \(T_2 = T_1(P_2/P_1)^{(\gamma-1)/\gamma} = 300(1/30)^{0.286}\).
\(30^{0.286} = e^{0.286\ln 30} = e^{0.973} = 2.65\). So T₂ = 300/2.65 = 113.4 K.
\(W_{ad} = \mu R(T_1 - T_2)/(\gamma -1) = 1\times 8.314 \times 186.6 / 0.4 = 3878\) J.
Adiabatic gives less work — the temperature drops, so the gas cools and pushes less.
(b) Adiabatic to same P₁/P₂ = 30: \(T_2 = T_1(P_2/P_1)^{(\gamma-1)/\gamma} = 300(1/30)^{0.286}\).
\(30^{0.286} = e^{0.286\ln 30} = e^{0.973} = 2.65\). So T₂ = 300/2.65 = 113.4 K.
\(W_{ad} = \mu R(T_1 - T_2)/(\gamma -1) = 1\times 8.314 \times 186.6 / 0.4 = 3878\) J.
Adiabatic gives less work — the temperature drops, so the gas cools and pushes less.
Example 11.11: Heat in an isochoric heating
5 L of nitrogen (diatomic, C_v = 5R/2) at 1 atm and 27 °C is heated at constant volume to 327 °C. Find the heat absorbed and the final pressure.
Number of moles: \(\mu = PV/RT = (10^5)(5\times 10^{-3})/(8.314\times 300) = 0.2004\) mol.
Q = µC_v ΔT = 0.2004 × (5/2)(8.314) × 300 = 1250 J.
Final P: \(P_2/P_1 = T_2/T_1 = 600/300 = 2\), so P₂ = 2 atm.
W = 0 (constant volume); ΔU = Q.
Q = µC_v ΔT = 0.2004 × (5/2)(8.314) × 300 = 1250 J.
Final P: \(P_2/P_1 = T_2/T_1 = 600/300 = 2\), so P₂ = 2 atm.
W = 0 (constant volume); ΔU = Q.
Interactive: Process Selector & Energy Calculator L3 Apply
Pick a process; set initial state and the final volume (or pressure). The applet computes W, Q and ΔU using ideal-gas formulas (γ = 1.4 for air; C_v = 5R/2).
Activity — Cloud-in-a-Bottle (Adiabatic Cooling)
L4 Analyse
Predict: Squeeze a plastic bottle containing a little water (and a puff of smoke) really hard; what happens when you suddenly release the squeeze?
- Take a 2-litre PET bottle. Put about 5 mL of water inside, swirl, then strike a match and drop in a tiny amount of smoke (just a wisp).
- Cap the bottle tightly. Squeeze hard with both hands and hold for a few seconds.
- Release the squeeze suddenly while looking at the inside.
Observation: A whitish cloud forms instantly inside the bottle the moment you release the squeeze.
Explanation: Releasing the squeeze causes the air to expand rapidly — almost adiabatically. Adiabatic expansion drops the temperature; the cool air can no longer hold all its water vapour, which condenses on the smoke nuclei to form tiny droplets — a cloud. Real clouds form when warm moist air rises and cools adiabatically.
Explanation: Releasing the squeeze causes the air to expand rapidly — almost adiabatically. Adiabatic expansion drops the temperature; the cool air can no longer hold all its water vapour, which condenses on the smoke nuclei to form tiny droplets — a cloud. Real clouds form when warm moist air rises and cools adiabatically.
Competency-Based Questions
A heat engine cycles 1 mol of an ideal monatomic gas (γ = 5/3, C_v = 12.47 J/mol·K) through: (A→B) isothermal expansion at 600 K from 1 L to 4 L, (B→C) isochoric cooling at 4 L from 600 K to 300 K, (C→A) isothermal compression at 300 K from 4 L to 1 L. R = 8.314 J/mol·K.
Q1. L1 Remember Define a quasi-static process.
An infinitely slow process during which the system is essentially in equilibrium at every instant — its state can be defined at every step and plotted on a P–V diagram.
Q2. L3 Apply Compute work in segment A→B.
\(W_{AB} = \mu RT \ln(V_2/V_1) = 1 \times 8.314 \times 600 \times \ln 4 = 4988 \times 1.386 = \boxed{6915\text{ J}}\). Heat in = same.
Q3. L3 Apply Compute work in segment C→A (isothermal compression at 300 K).
\(W_{CA} = \mu RT \ln(V_A/V_C) = 1 \times 8.314 \times 300 \times \ln(1/4) = -3458\text{ J}\). Negative — work done on gas.
Q4. L3 Apply Compute net work in one full cycle.
W_BC = 0 (isochoric). Total W = 6915 + 0 + (−3458) = 3457 J. Heat in (only at T = 600 K, A→B) = 6915 J. Efficiency η = 3457/6915 ≈ 50 % (close to Carnot for these temperatures).
Q5. L5 Evaluate Why is an adiabatic expansion always cooler than the isothermal between the same volumes?
In an adiabatic expansion, no heat enters from outside; the work done by the gas comes purely at the cost of internal energy, so T must drop. In an isothermal expansion, the gas absorbs the same amount of heat from a reservoir as the work it does, so T stays constant. Hence the adiabatic curve always sits below the isotherm at the same V (steeper drop).
Assertion-Reason Questions
Assertion (A): The slope of an adiabatic curve on a P–V diagram is steeper than that of an isothermal at the same point.
Reason (R): The slope of an adiabat is −γ P/V, while the slope of an isotherm is −P/V; γ > 1.
Answer: A. Differentiate \(PV = const\) (isotherm) and \(PV^\gamma = const\) (adiabat) to verify the slope formulas. γ > 1 makes the adiabat steeper.
Assertion (A): In a free expansion of an ideal gas, the temperature does not change.
Reason (R): Free expansion has Q = 0 and W = 0, so ΔU = 0; for an ideal gas, U depends only on T, hence T is constant.
Answer: A. Free expansion is the textbook proof that ideal-gas U depends only on T.
Assertion (A): An isothermal expansion of an ideal gas requires heat absorbed from the surroundings.
Reason (R): ΔU = 0 in an isothermal process (ideal gas), so by First Law Q = W, and W > 0 in expansion implies Q > 0.
Answer: A. Without heat input, T would drop on expansion. Heat input compensates the cooling.
Frequently Asked Questions - Thermodynamic Processes
What is the main concept covered in Thermodynamic Processes?
In NCERT Class 11 Physics Chapter 11 (Thermodynamics), "Thermodynamic Processes" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Thermodynamic Processes useful in real-life applications?
Real-life applications of Thermodynamic Processes from NCERT Class 11 Physics Chapter 11 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Thermodynamic Processes?
Key formulas in Thermodynamic Processes (NCERT Class 11 Physics Chapter 11 Thermodynamics) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 11?
NCERT Class 11 Physics Chapter 11 (Thermodynamics) is structured so each part builds on the previous one. Thermodynamic Processes connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Thermodynamic Processes?
CBSE board questions from Thermodynamic Processes typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Thermodynamic Processes lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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