This MCQ module is based on: First Law
TOPIC 15 OF 33
First Law
🎓 Class 11
Physics
CBSE
Theory
Ch 11 – Thermodynamics
⏱ ~14 min
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First Law
11.7 The First Law of Thermodynamics
From experience, the change in internal energy of a system can be brought about in two distinct ways:
- By heating — putting it in contact with a hotter body (heat absorbed Q).
- By doing work on it — compressing a gas, stirring a liquid, rubbing surfaces together.
Joule's classic 1840s experiments showed that whether you heat water with a flame or stir it with a paddle, the rise in temperature is the same provided the energy delivered is the same — heat and mechanical work are interchangeable forms of energy. This idea, made quantitative, is the celebrated First Law.
First Law of Thermodynamics:
\[\boxed{\;\Delta U = Q - W\;}\]
where ΔU = change in internal energy, Q = heat absorbed by the system, W = work done by the system on the surroundings. The First Law is simply conservation of energy, generalised to include heat as a form of energy.
11.7.1 Differential Form & Sign Conventions
For an infinitesimal process: \(dU = \delta Q - \delta W\).
Sign rules (NCERT convention):
Sign rules (NCERT convention):
- \(Q > 0\) — heat added to the system.
- \(W > 0\) — work done by the system on surroundings.
- \(W < 0\) — work done on the system (compression).
- The sign of \(\Delta U\) is fixed by Q and W.
Watch the chemists' convention: Many chemistry textbooks (and IUPAC) write the First Law as ΔU = Q + W, using W = work done on the system. The physics is identical; just stick to one convention. NCERT physics uses ΔU = Q − W.
11.7.2 Work Done by a Gas — General Formula
For a quasi-static (slow, reversible) volume change from \(V_1\) to \(V_2\):
\[W = \int_{V_1}^{V_2} P\, dV\]
The integral equals the area under the P–V curve. Different paths between the same endpoints give different values of W (and Q), even though ΔU is the same.
11.7.3 Three Special Cases at a Glance
| Process | Constraint | W | Q | ΔU |
|---|---|---|---|---|
| Free expansion | insulated, into vacuum | 0 | 0 | 0 |
| Cyclic process | system returns to initial state | = ∮P dV = area enclosed | = W | 0 |
| Isolated system | no heat or work exchange | 0 | 0 | 0 |
Note on cyclic processes: Since U is a state function, going around a closed loop in the P–V plane gives \(\Delta U_\text{cycle} = 0\), hence \(Q_\text{cycle} = W_\text{cycle}\). All the heat absorbed in one cycle is converted to work — this is the basic principle of any heat engine (next part).
11.8 Specific Heat Capacities of an Ideal Gas
For a gas, the heat needed to raise temperature depends on whether the volume is held fixed or the pressure is held fixed. Hence two molar specific heats:
At constant volume: \(\delta Q = \mu C_v\, dT\). Since W = 0, \(\delta Q = dU\), so:
\[C_v = \frac{1}{\mu}\frac{dU}{dT}\]
At constant pressure: \(\delta Q = \mu C_p\, dT\). Now W = P dV = µR dT (using PV = µRT at constant P), so:
\[\mu C_p\, dT = dU + \mu R\, dT \;\Rightarrow\; C_p - C_v = R \quad \text{(Mayer's relation)}\]
| Gas type | Degrees of freedom f | C_v | C_p | γ = C_p/C_v |
|---|---|---|---|---|
| Monatomic (He, Ar, Ne) | 3 | (3/2)R = 12.5 | (5/2)R = 20.8 | 5/3 ≈ 1.67 |
| Diatomic (H₂, N₂, O₂) at room T | 5 | (5/2)R = 20.8 | (7/2)R = 29.1 | 7/5 = 1.40 |
| Polyatomic (NH₃, CH₄) | 6+ | 3R ≈ 24.9 | 4R ≈ 33.2 | 4/3 ≈ 1.33 |
(C_v and C_p in J mol⁻¹ K⁻¹.) γ is also called the adiabatic exponent; it appears in many thermodynamic formulas and is closely related to the speed of sound in a gas.
Worked Examples
Example 11.4: Apply the First Law
A system absorbs 250 J of heat and does 100 J of work on the surroundings. (a) Find the change in internal energy. (b) Repeat if instead 100 J of work is done on the system while 250 J of heat is absorbed.
(a) ΔU = Q − W = 250 − 100 = 150 J (gain).
(b) Now W = −100 J (work done on system). ΔU = 250 − (−100) = 350 J.
The same heat input plus external work gives a much larger ΔU.
(b) Now W = −100 J (work done on system). ΔU = 250 − (−100) = 350 J.
The same heat input plus external work gives a much larger ΔU.
Example 11.5: Cyclic process — work and heat
An ideal gas goes through a cyclic process A → B → C → A in the P–V plane. Heat absorbed in A→B = 800 J, in B→C = −500 J (released), in C→A = 0 (adiabatic). Find the net work done by the gas in one cycle and the efficiency if Q_h = 800 J is the only heat input.
For a cycle, ΔU = 0, so W_cycle = Q_cycle = 800 + (−500) + 0 = 300 J.
Efficiency η = W/Q_h = 300/800 = 0.375 = 37.5 %.
This is the kind of energy bookkeeping that defines a heat engine.
Efficiency η = W/Q_h = 300/800 = 0.375 = 37.5 %.
This is the kind of energy bookkeeping that defines a heat engine.
Example 11.6: Constant-pressure heating of a gas
2.0 moles of an ideal monatomic gas at 27 °C is heated at constant pressure until its temperature reaches 127 °C. Find (i) the heat absorbed, (ii) the work done by the gas, (iii) the change in internal energy. R = 8.314 J/mol·K.
ΔT = 100 K. C_v = 3R/2 = 12.47, C_p = 5R/2 = 20.79 (J/mol·K).
(i) Q = µC_p ΔT = 2 × 20.79 × 100 = 4158 J.
(ii) W = P ΔV = µR ΔT = 2 × 8.314 × 100 = 1663 J.
(iii) ΔU = Q − W = 4158 − 1663 = 2495 J. Check: µ C_v ΔT = 2 × 12.47 × 100 = 2495 J ✓.
(i) Q = µC_p ΔT = 2 × 20.79 × 100 = 4158 J.
(ii) W = P ΔV = µR ΔT = 2 × 8.314 × 100 = 1663 J.
(iii) ΔU = Q − W = 4158 − 1663 = 2495 J. Check: µ C_v ΔT = 2 × 12.47 × 100 = 2495 J ✓.
Example 11.7: Joule's mechanical equivalent of heat
A bullet of mass 50 g moving at 200 m/s embeds in a wooden block. Assuming all the kinetic energy goes into heat shared equally between bullet and block, and that the block weighs 1 kg with specific heat 1700 J/kg·K, while the bullet (lead) has s = 130 J/kg·K, find the temperature rise of the bullet.
KE = ½ m v² = ½ × 0.050 × 200² = 1000 J. Half goes to bullet: 500 J.
ΔT_bullet = 500 / (0.050 × 130) = 76.9 K. Lead has very low specific heat — bullet heats up appreciably.
ΔT_bullet = 500 / (0.050 × 130) = 76.9 K. Lead has very low specific heat — bullet heats up appreciably.
Interactive: First-Law Calculator (any 2 → 3rd) L3 Apply
Enter any two of Q, W, ΔU; the applet computes the missing one and indicates the dominant energy flow.
Activity — Bicycle-Pump Adiabatic Heating
L4 Analyse
Predict: Pump up a flat bicycle tyre vigorously and immediately touch the metal nozzle near the valve. What do you feel?
- Take a hand-held bicycle pump (with a metal cylinder).
- Pump rapidly to inflate a tyre or empty container with high resistance.
- After 30 seconds of vigorous pumping, touch the cylinder near the bottom (close to the valve).
Observation: The cylinder near the valve is noticeably warm, sometimes hot to touch.
Explanation: Rapid compression of air is approximately adiabatic (Q ≈ 0 because there's no time for heat exchange). The First Law gives ΔU = −W. Since W < 0 (you do work on the gas), ΔU > 0 — internal energy and hence temperature rise. Diesel engines exploit exactly this effect to ignite fuel without spark plugs.
Explanation: Rapid compression of air is approximately adiabatic (Q ≈ 0 because there's no time for heat exchange). The First Law gives ΔU = −W. Since W < 0 (you do work on the gas), ΔU > 0 — internal energy and hence temperature rise. Diesel engines exploit exactly this effect to ignite fuel without spark plugs.
Competency-Based Questions
A small high-pressure gas cylinder contains 0.5 mol of helium (monatomic, C_v = 12.5 J/mol·K). 1500 J of heat is supplied and the gas does 600 J of work pushing back a frictionless piston. Take initial temperature 300 K.
Q1. L1 Remember State the First Law of Thermodynamics in equation form, with sign conventions used.
ΔU = Q − W. Q is positive when heat is added to the system; W is positive when work is done by the system on the surroundings.
Q2. L3 Apply Compute the change in internal energy.
ΔU = 1500 − 600 = 900 J.
Q3. L3 Apply Calculate the rise in temperature of the helium.
ΔT = ΔU/(µC_v) = 900/(0.5 × 12.5) = 144 K. Final T = 300 + 144 = 444 K.
Q4. L4 Analyse What fraction of the supplied heat went into useful work?
Fraction = W/Q = 600/1500 = 0.40 = 40 %. The remaining 60% raised internal energy. (For a real heat engine, the second law caps this fraction further.)
Q5. L5 Evaluate Why is it physically impossible to construct a "perpetual-motion machine of the first kind" — one that produces continuous work without any energy input?
Because the First Law (ΔU = Q − W) demands energy conservation: in a cycle ΔU = 0 so W = Q. Without heat input (Q = 0), no net work can come out (W = 0). A "perpetual" machine that produced energy from nothing would violate the conservation of energy itself, the deepest empirical law in physics.
Assertion-Reason Questions
Assertion (A): The internal energy of a system can change without any heat being added.
Reason (R): The First Law states ΔU = Q − W, so even with Q = 0, work done on the system can change ΔU.
Answer: A. Adiabatic compression (Q = 0) raises U through W < 0. R correctly explains A.
Assertion (A): In a cyclic process, the heat absorbed equals the work done by the system.
Reason (R): Internal energy is a state function, so ΔU = 0 over a complete cycle.
Answer: A. First Law on a cycle: 0 = Q_cycle − W_cycle, so Q_cycle = W_cycle. R is exactly why.
Assertion (A): For an ideal gas, C_p > C_v always.
Reason (R): At constant pressure the gas does extra work on the surroundings during expansion, requiring extra heat input.
Answer: A. C_p − C_v = R > 0 (Mayer's relation), exactly because of the work term R describes.
Frequently Asked Questions - First Law
What is the main concept covered in First Law?
In NCERT Class 11 Physics Chapter 11 (Thermodynamics), "First Law" 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 First Law useful in real-life applications?
Real-life applications of First Law 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 First Law?
Key formulas in First Law (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. First Law 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 First Law?
CBSE board questions from First Law 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 First Law 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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