First Law of Thermodynamics
Class 11 Physics notes covering First Law, energy conservation, sign convention, cyclic processes, numericals and PYQs.
The First Law of Thermodynamics is the energy-accounting law of physics. It connects heat supplied to a system, work done by the system and change in internal energy. Once this law is clear, every thermodynamics process becomes a balance sheet: energy entering the system either remains stored as internal energy or leaves as work output.
Introduction
The First Law of Thermodynamics is the principle of conservation of energy applied to heat, work and internal energy.
In mechanics, conservation of energy often appears as conversion between kinetic energy and potential energy. In thermodynamics, energy can cross the boundary of a system as heat or work, and the energy stored inside the system is called internal energy. The First Law tells us that these quantities cannot be chosen independently. If heat enters a gas and the gas expands, some part of the incoming energy may increase molecular energy and some part may push the piston outward.
A top student treats every problem as a system-boundary question. First choose the system. Then ask whether heat enters or leaves. Next ask whether work is done by the system or on the system. Finally calculate how internal energy changes. This discipline prevents almost every sign-convention mistake.
Heat Input
Energy enters because of temperature difference. It is written as positive when supplied to the system.
Work Output
Energy leaves as mechanical work when the system expands and pushes surroundings.
Internal Storage
Energy remains in the microscopic motion and interaction of molecules as internal energy.
First Law of Thermodynamics
The mathematical statement is ΔQ = ΔU + ΔW, where heat supplied equals increase in internal energy plus work done by the system.
Meaning of Each Term
ΔQ
Heat exchanged with the system. It is positive when heat is supplied to the system and negative when heat is rejected by the system.
ΔU
Change in internal energy. It measures change in microscopic energy of molecules due to random motion and intermolecular interactions.
ΔW
Work done by the system on surroundings. For expansion of gas, it is usually positive in the convention used here.
Physical Interpretation
If you supply heat to gas in a cylinder, the gas may become hotter and it may also expand. Becoming hotter means internal energy increases. Expanding means gas does work on the piston. The First Law says the heat supplied must be equal to the total of these two energy effects.
Microscopic Interpretation
At molecular level, heat supplied increases molecular kinetic energy, changes molecular separation, or both. When a gas expands, molecules transfer momentum to the moving piston and energy leaves the gas as work. Therefore, the law is a bridge between microscopic molecular energy and macroscopic mechanical work.
Derivation Idea
The First Law is not derived from a more basic thermodynamic equation at Class 11 level; it is the conservation of energy written for thermodynamic systems. If energy ΔQ enters as heat and ΔW leaves as work, the remaining energy must appear as increase in internal energy. Hence ΔU = ΔQ - ΔW, or equivalently ΔQ = ΔU + ΔW.
Core Diagrams
These NCERT-style diagrams use black outlines and red arrows to show energy flow, work direction and P-V graph interpretation.
System Receiving Heat
System Doing Work
Expansion Process
Compression Process
Energy Flow Diagram
Cyclic Process Loop
P-V Diagram
Constant Volume Process
Constant Pressure Process
Sign Convention
Sign convention is the most common source of errors in First Law questions. The physics convention used here is: heat supplied to the system is positive and work done by the system is positive.
| Case | Physical meaning | Sign | First Law effect | Memory trick |
|---|---|---|---|---|
| Heat supplied to system | Energy enters as heat | ΔQ positive | Tends to increase ΔU or produce work | Heat in is plus |
| Heat rejected by system | Energy leaves as heat | ΔQ negative | Tends to decrease ΔU | Heat out is minus |
| Work done by system | Expansion pushes surroundings | ΔW positive | Subtracts from internal energy in ΔU = Q - W | Output work is plus W |
| Work done on system | Compression by surroundings | ΔW negative for work by system | Increases internal energy if no heat escapes | Compression makes W by gas minus |
Separate Sign Tables
Heat Supplied
When a burner heats gas or a hot reservoir gives energy to the system, ΔQ is positive. Example: Q = +600 J.
Heat Rejected
When the system cools or gives heat to surroundings, ΔQ is negative. Example: Q = -250 J.
Work Done by System
Expansion work by gas is positive. Example: a piston moves outward and W = +400 J.
Work Done on System
Compression means surroundings do work on gas. Work by gas is negative. Example: W = -400 J.
Exam Traps
- Do not change convention in the middle of the solution.
- Do not assume heat supplied always increases internal energy; work may carry energy away.
- Do not assume compression has positive work by gas; it is negative in this convention.
- In a cyclic process, internal energy change is zero even if heat and work are nonzero.
Energy Conservation Principle
The First Law is a strict energy balance: energy cannot appear from nowhere and cannot disappear into nothing.
For a system, heat and work are transfer modes across the boundary. Internal energy is the energy stored inside. If more energy enters than leaves, internal energy increases. If more energy leaves than enters, internal energy decreases. If energy entering as heat equals energy leaving as work, internal energy remains constant.
Heat supplied equals increase in internal energy plus work done by system.
Internal energy increases when heat input exceeds work output.
Work done by a gas equals area under the P-V curve.
For a cyclic process, final state equals initial state.
For a complete cycle, net heat equals net work.
Real-life interpretation is simple: if you heat a pressure cooker, energy enters. Some energy increases internal energy of water and steam; if steam pushes out through the valve, energy also leaves as flow and work. In a car engine, fuel energy becomes heat, part of it becomes work, and the rest is rejected to surroundings. The First Law is the reason energy accounting is possible in engines, refrigerators, turbines and biological systems.
Applications of First Law
Different processes simplify the First Law in different ways. Recognising the process quickly is a scoring skill.
| Application | Condition | First Law Result | Physical Meaning | Exam Clue |
|---|---|---|---|---|
| Constant volume process | ΔV = 0 | W = 0, so Q = ΔU | Heat changes internal energy only. | Rigid container, vertical P-V graph. |
| Constant pressure process | P constant | Q = ΔU + PΔV | Heat both raises internal energy and does work. | Horizontal P-V graph. |
| Expansion process | ΔV positive | W positive | System gives energy to surroundings as work. | Piston moves outward. |
| Compression process | ΔV negative | W by system negative | Surroundings give energy to system as work. | Piston moves inward. |
| Heating process | Q positive | ΔU depends on W | Supplied heat may increase U and/or produce work. | Burner, hot reservoir. |
| Cooling process | Q negative | Internal energy often decreases | Energy leaves as heat. | System in colder surroundings. |
| Cyclic process | Final state = initial state | ΔU = 0, Q = W | Net heat becomes net work over cycle. | Closed P-V loop. |
Conceptual Examples
Gas in Rigid Cylinder
Volume is fixed, so no boundary work occurs. Heat supplied increases internal energy and temperature.
Gas Under Movable Piston
Heat supplied can expand the gas. Energy divides between internal energy rise and work output.
Rapid Compression
Heat exchange is small, so work done on gas mostly increases internal energy and temperature.
Heat Engine Cycle
After each cycle the working substance returns to its original state, so net internal energy change is zero.
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Cyclic Process
A cyclic process is a thermodynamic process in which the system returns to its initial state after a series of changes.
Because the final state is the same as the initial state, every state property returns to its original value. Internal energy is a state property, so the net change in internal energy over a complete cycle is zero. This is the key idea behind heat engines and refrigerators.
On a P-V diagram, a cyclic process appears as a closed loop. The area enclosed by the loop gives net work. A clockwise loop represents positive work done by the system, like a heat engine. An anticlockwise loop represents net work done on the system, like a refrigerator or heat pump.
Important PYQ Observations
- In a cycle, do not put Q = 0. Put ΔU = 0.
- Net heat and net work over a cycle are equal in magnitude and sign under the convention used here.
- The working substance may absorb heat in one part and reject heat in another part.
- Efficiency questions use net work output divided by heat absorbed from hot reservoir.
Common Mistakes and Memory Tricks
Mistake: Q = 0 in a cycle
Wrong. In a cycle, ΔU = 0. Net heat generally equals net work and may be nonzero.
Mistake: Work is always positive
Expansion work by gas is positive. Compression work by gas is negative.
Mistake: Heat always raises temperature
Heat can become work or can be involved in phase change. Internal energy rise depends on Q - W.
Memory Trick
Heat in fills the system. Work out empties it. So ΔU = heat in - work out.
50 Solved Numericals
Solutions are hidden so students can attempt first and then reveal the full method.
Show Solution
Given: Q = +500 J, W = +180 J.
Formula: ΔU = Q - W.
Solution: ΔU = 500 - 180 = 320 J.
Final Answer: Internal energy increases by 320 J.
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Given: Adiabatic process: Q = 0. Work done on gas = 250 J, so W by gas = -250 J.
Formula: ΔU = Q - W.
Solution: ΔU = 0 - (-250) = 250 J.
Final Answer: Internal energy increases by 250 J.
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Given: Q = -120 J, W = +80 J.
Formula: ΔU = Q - W.
Solution: ΔU = -120 - 80 = -200 J.
Final Answer: Internal energy decreases by 200 J.
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Given: ΔU = 300 J, W = 100 J.
Formula: Q = ΔU + W.
Solution: Q = 300 + 100 = 400 J.
Final Answer: Heat supplied = 400 J.
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Given: Cyclic process: ΔU = 0, W = 700 J.
Formula: Q = W for a cycle.
Solution: Q = 700 J.
Final Answer: Net heat absorbed = 700 J.
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Given: Constant volume: ΔV = 0, Q = 900 J.
Formula: W = 0, ΔU = Q - W.
Solution: W = 0 and ΔU = 900 J.
Final Answer: Work = 0, internal energy increases by 900 J.
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Given: P = 1 × 105 Pa, ΔV = 3 L = 3 × 10-3 m3.
Formula: W = PΔV.
Solution: W = 1 × 105 × 3 × 10-3 = 300 J.
Final Answer: Work done by gas = 300 J.
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Given: Q = 1000 J, ΔU = 650 J.
Formula: W = Q - ΔU.
Solution: W = 1000 - 650 = 350 J.
Final Answer: Work done by system = 350 J.
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Given: Q = -400 J, W by contents = -150 J.
Formula: ΔU = Q - W.
Solution: ΔU = -400 - (-150) = -250 J.
Final Answer: Internal energy decreases by 250 J.
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Given: W = 0, ΔU = -200 J.
Formula: Q = ΔU + W.
Solution: Q = -200 + 0 = -200 J.
Final Answer: Heat rejected = 200 J.
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Given: Q = 600 J, ΔU = 250 J.
Formula: ΔU = Q - W.
Solution: W = Q - ΔU = 600 - 250 = 350 J.
Final Answer: Work done = 350 J.
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Given: Q = -100 J, W by gas = -500 J.
Formula: ΔU = Q - W.
Solution: ΔU = -100 - (-500) = 400 J.
Final Answer: Internal energy increases by 400 J.
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Given: W = 220 J, ΔU = -50 J.
Formula: Q = ΔU + W.
Solution: Q = -50 + 220 = 170 J.
Final Answer: Heat supplied = 170 J.
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Given: Constant volume means no boundary displacement.
Formula: W = PΔV = 0, ΔU = Q.
Solution: ΔU = 450 J.
Final Answer: All 450 J increases internal energy.
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Given: P = 2 × 105 Pa, ΔV = 0.004 m3.
Formula: W = PΔV.
Solution: W = 2 × 105 × 0.004 = 800 J.
Final Answer: Work done by gas = 800 J.
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Given: Q = +300 J, W by system = -200 J.
Formula: ΔU = Q - W.
Solution: ΔU = 300 - (-200) = 500 J.
Final Answer: Internal energy increases by 500 J.
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Given: Qnet = 1200 - 800 = 400 J, ΔU = 0.
Formula: For cycle, Qnet = Wnet.
Solution: Wnet = 400 J.
Final Answer: Net work output = 400 J.
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Given: Q = 1000 J, W = 350 J.
Formula: ΔU = Q - W.
Solution: ΔU = 1000 - 350 = 650 J.
Final Answer: Internal energy rises by 650 J.
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Given: ΔU = -600 J, W = 100 J.
Formula: Q = ΔU + W.
Solution: Q = -600 + 100 = -500 J.
Final Answer: Heat rejected = 500 J.
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Given: V1 = 10 L, V2 = 6 L, ΔV = -4 × 10-3 m3.
Formula: W = PΔV.
Solution: W = 1 × 105 × (-4 × 10-3) = -400 J.
Final Answer: Work by gas = -400 J; work on gas = 400 J.
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Given: Isothermal ideal gas: ΔU = 0, Q = 500 J.
Formula: ΔU = Q - W.
Solution: 0 = 500 - W, so W = 500 J.
Final Answer: ΔU = 0, W = 500 J.
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Given: Adiabatic: Q = 0, W = 300 J.
Formula: ΔU = Q - W.
Solution: ΔU = -300 J.
Final Answer: Internal energy decreases by 300 J.
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Given: Q = +700 J, W by gas = -400 J.
Formula: ΔU = Q - W.
Solution: ΔU = 700 - (-400) = 1100 J.
Final Answer: Internal energy increases by 1100 J.
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Given: Cycle: ΔU = 0, Q = -200 J.
Formula: Q = W.
Solution: W = -200 J.
Final Answer: Net work is done on system; work by system = -200 J.
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Given: ΔV = 2 × 10-3 m3, P = 5 × 105 Pa, Q = 1500 J.
Formula: W = PΔV, ΔU = Q - W.
Solution: W = 1000 J. ΔU = 1500 - 1000 = 500 J.
Final Answer: Internal energy increases by 500 J.
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Given: ΔU = 0, W = -250 J.
Formula: Q = ΔU + W.
Solution: Q = -250 J. Negative W means compression.
Final Answer: Heat rejected = 250 J; compression occurs.
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Given: Constant volume, heating.
Formula: W = 0, ΔU = Q.
Solution: Heat supplied is positive, so ΔU positive.
Final Answer: W = 0, internal energy increases.
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Given: Isothermal ideal gas: ΔU = 0. W by gas = -600 J.
Formula: Q = ΔU + W.
Solution: Q = 0 - 600 = -600 J.
Final Answer: Heat rejected = 600 J.
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Given: Q = W = 900 J.
Formula: ΔU = Q - W.
Solution: ΔU = 0. For ideal gas, temperature is constant.
Final Answer: Ideal gas process is isothermal if only temperature decides U.
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Given: ΔU = -100 J, Q = -300 J.
Formula: W = Q - ΔU.
Solution: W = -300 - (-100) = -200 J.
Final Answer: Work by system = -200 J; 200 J work done on it.
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Given: P1 = 1 × 105 Pa, P2 = 3 × 105 Pa, ΔV = 4 × 10-3 m3.
Formula: W = PavgΔV.
Solution: Pavg = 2 × 105 Pa. W = 800 J.
Final Answer: Work = 800 J.
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Given: ΔP = 3 × 105 Pa, ΔV = 4 × 10-3 m3.
Formula: Net work = area of rectangle.
Solution: W = 3 × 105 × 4 × 10-3 = 1200 J.
Final Answer: Net work = +1200 J.
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Given: Cyclic process: ΔU = 0, Q = 1500 J.
Formula: Q = W.
Solution: W = 1500 J.
Final Answer: Net work = 1500 J, ΔU = 0.
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Given: Q = -500 J, ΔV = 8 × 10-3 m3, P = 2.5 × 105 Pa.
Formula: W = PΔV, ΔU = Q - W.
Solution: W = 2000 J. ΔU = -500 - 2000 = -2500 J.
Final Answer: Internal energy decreases by 2500 J.
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Given: Same initial and final states.
Formula: Internal energy is a state function.
Solution: ΔU is independent of path.
Final Answer: Change in internal energy = 400 J.
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Given: Q = 2000 J, W = 750 J.
Formula: ΔU = Q - W.
Solution: ΔU = 1250 J.
Final Answer: Internal energy increases by 1250 J.
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Given: Vertical line means volume constant.
Formula: W = area under P-V curve = 0 because ΔV = 0.
Solution: No boundary work.
Final Answer: W = 0.
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Given: Constant pressure, ΔV = 3 L = 3 × 10-3 m3.
Formula: W = PΔV.
Solution: W = 3 × 105 × 3 × 10-3 = 900 J.
Final Answer: Work = 900 J.
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Given: Q = 0, work input = 1000 J, so W by gas = -1000 J.
Formula: ΔU = Q - W.
Solution: ΔU = 0 - (-1000) = 1000 J.
Final Answer: Q = 0, W = -1000 J, ΔU = +1000 J.
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Given: Q = -400 J, ΔU = -900 J.
Formula: W = Q - ΔU.
Solution: W = -400 - (-900) = 500 J.
Final Answer: Work done by gas = 500 J.
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Given: Path 1: Q1 = 900 J, W1 = 500 J.
Formula: ΔU = Q - W and same between A and B.
Solution: ΔU = 900 - 500 = 400 J. For path 2, Q2 = ΔU + W2 = 400 + 300 = 700 J.
Final Answer: Heat along path 2 = 700 J.
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Given: Qin = 2000 J, Qout = 1400 J.
Formula: W = Qin - Qout, efficiency = W/Qin.
Solution: W = 600 J. Efficiency = 600/2000 = 0.30.
Final Answer: Work = 600 J; efficiency = 30%.
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Given: For complete cycle, ΔU = 0. Qnet = 500 - 200 = 300 J.
Formula: Wnet = Qnet.
Solution: Wnet = 300 J.
Final Answer: Net work by system = 300 J.
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Given: P = aV, a = 2 × 108, V1 = 10-3 m3, V2 = 3 × 10-3 m3.
Formula: W = ∫PdV = ∫aVdV = a(V22 - V12)/2.
Solution: W = 2 × 108[(9 - 1)×10-6]/2 = 800 J.
Final Answer: Work = 800 J.
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Given: Q = 1200 J, Wtotal = 300 + 200 = 500 J.
Formula: ΔU = Q - W.
Solution: ΔU = 1200 - 500 = 700 J.
Final Answer: Internal energy increases by 700 J.
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Given: Step 1: Q1 = 700 J, W1 = 400 J. Step 2: Q2 = -100 J, W2 = -250 J.
Formula: ΔU = Q - W for each step.
Solution: ΔU1 = 300 J. ΔU2 = -100 - (-250) = 150 J. Total = 450 J.
Final Answer: Total internal energy increases by 450 J.
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Given: Cycle: ΔU = 0. Net W = 500 - 200 + 100 = 400 J.
Formula: Qnet = Wnet.
Solution: Qnet = 400 J.
Final Answer: Net heat absorbed = 400 J.
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Given: ΔU = -300 J, W by system = -700 J.
Formula: Q = ΔU + W.
Solution: Q = -300 - 700 = -1000 J.
Final Answer: Heat rejected = 1000 J.
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Given: Pavg = 4 × 105 Pa, ΔV = 6 × 10-3 m3, Q = 3000 J.
Formula: W = PavgΔV, ΔU = Q - W.
Solution: W = 2400 J. ΔU = 3000 - 2400 = 600 J.
Final Answer: Internal energy increases by 600 J.
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Given: Cycle: ΔU = 0. W = 900 J, Qout = 2100 J.
Formula: Qin - Qout = W.
Solution: Qin = 900 + 2100 = 3000 J.
Final Answer: Heat absorbed = 3000 J.
50 PYQs and Exam Questions
Answers are hidden. Click only after attempting the question.
Show Answer
Explanation: The First Law states that heat supplied to a system is used to increase internal energy and to do work by the system.
ΔQ = ΔU + ΔW.
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Explanation: It says energy is neither created nor destroyed in a thermodynamic process; it only changes form between heat, work and internal energy.
It is conservation of energy applied to thermodynamics.
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Explanation: ΔU represents change in internal energy of the system, the microscopic energy stored in molecular motion and interactions.
It is a state-function change.
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Explanation: At constant volume, boundary displacement is zero, so P-V work is zero.
W = 0.
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Explanation: In a cycle, the system returns to its initial state, so ΔU = 0.
Q = W.
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Explanation: Internal energy is a state function and final state equals initial state.
Answer: (C) ΔU.
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Explanation: Using ΔU = Q - W = 100 - 40 = 60 J.
Answer: (B) 60 J.
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Explanation: Adiabatic means no heat exchange.
Answer: (A) Q = 0.
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Explanation: Isochoric means constant volume, so W = 0 and ΔU = Q.
Answer: (B) Change in internal energy.
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Explanation: Q = 0 and W by gas is negative; therefore ΔU = -W is positive.
Answer: (B) Increases.
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Explanation: The area under the P-V curve gives ∫PdV, the work done by the gas in a quasi-static process.
Answer: Work done.
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Explanation: The initial and final states are the same, and internal energy depends only on state.
Answer: State function returns to original value.
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Explanation: A clockwise cycle encloses positive area for work done by the system.
Answer: Positive work.
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Explanation: Only internal energy change depends only on endpoints.
Answer: ΔU.
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Explanation: At constant pressure, the area under the P-V graph is a rectangle.
Answer: W = P(V2 - V1).
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Explanation: For an ideal gas isothermal expansion, ΔU = 0, so Q = W for that process. Broader engine-cycle limitations come from the second law.
Answer: In a process yes under suitable conditions; not as a complete cyclic engine with single reservoir.
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Explanation: ΔU is a state-function change, so it is identical for both paths.
Answer: Same for both paths.
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Explanation: External pressure is zero, so W = PextΔV = 0.
Answer: No opposing pressure, no work output.
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Explanation: It does work without receiving heat, so internal energy decreases. For ideal gas, lower internal energy means lower temperature.
Answer: Work is done at the cost of internal energy.
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Explanation: The enclosed area equals net work done per cycle.
Answer: Net work output if clockwise.
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Explanation: Heating and working are transfers across the system boundary, while internal energy is the system's microscopic energy store.
Answer: Energy input by heating minus work output changes internal energy.
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Explanation: They depend on the route taken between states, not only on initial and final equilibrium states.
Answer: They are process quantities.
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Explanation: It is energy transferred from the gas to surroundings when the gas expands against external pressure.
Answer: Boundary work during expansion.
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Explanation: Rigid walls make volume constant, so boundary work is zero and supplied heat equals internal energy change.
Answer: W = 0, so ΔU = Q.
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Explanation: The system returns to its starting state, so internal energy returns to its original value.
Answer: Net ΔU = 0.
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Explanation: With work done by system positive, internal energy change equals heat supplied minus work done.
Answer: ΔU = Q - W.
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Explanation: Positive heat means heat is supplied to the system from surroundings.
Answer: Energy enters as heat.
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Explanation: Negative work by a gas means the gas is compressed and work is done on it.
Answer: Compression work.
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Explanation: Quick compression is approximately adiabatic; work done on gas increases internal energy.
Answer: Work input raises internal energy.
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Explanation: Yes, in adiabatic expansion a gas can do work using its internal energy.
Answer: Yes.
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Explanation: The gas transfers energy mechanically to the surroundings.
Answer: Work.
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Explanation: Since no boundary work is done, supplied heat increases internal energy.
Answer: Internal energy increases.
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Explanation: Total energy is conserved; heat, work and internal energy changes balance.
Answer: Energy is transformed, not destroyed.
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Explanation: Boundary work is zero because W = PΔV.
Answer: W = 0.
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Explanation: During expansion, gas does work; if heat input is insufficient, internal energy and temperature decrease.
Answer: Work output can reduce internal energy.
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Explanation: For a quasi-static process, work is the integral of pressure with respect to volume.
Answer: W = ∫PdV.
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Explanation: The First Law is energy conservation only; direction and feasibility require the Second Law.
Answer: It balances energy but does not set process direction.
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Explanation: Internal energy is a property of state; heat is a transfer during a process and changes with path.
Answer: State function versus process quantity.
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Explanation: Temperature is constant, so internal energy change is zero. From the First Law, Q = W.
Answer: Supplied heat becomes boundary work.
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Explanation: Q = 0 and work is done on gas, so W by gas is negative and ΔU is positive.
Answer: Internal energy increases.
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Explanation: Both statements are true and the reason correctly explains the assertion.
Answer: Both true; reason is correct.
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Explanation: Reason is true, but assertion is false because supplied heat can be used fully or partly for work.
Answer: Assertion false, reason true.
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Explanation: Both are true and the reason correctly explains the assertion.
Answer: Both true; reason is correct.
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Explanation: Both statements are true and the reason explains the assertion.
Answer: Both true; reason is correct.
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Explanation: Both assertion and reason are false in this context; heat and work are path dependent process quantities.
Answer: Both false.
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Explanation: Heat is positive, work by gas is positive, and internal energy change depends on Q - W.
Answer: Q > 0, W > 0, ΔU = Q - W.
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Explanation: Boundary work is nearly zero, so heat mainly increases internal energy of contents and raises temperature/pressure.
Answer: Approximately ΔU = Q.
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Explanation: Insulated means Q = 0; compression means W by gas is negative; therefore ΔU increases.
Answer: ΔU positive.
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Explanation: Since ΔU = 0 over a cycle, net heat absorbed equals net work output.
Answer: Qnet = Wnet.
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Explanation: Insulated gives Q = 0. Vacuum gives W = 0. Therefore ΔU = 0.
Answer: Q = 0, W = 0, ΔU = 0.
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Explanation: Yes. If enough work is done on the system, internal energy can increase despite heat loss.
Answer: Yes, work input can dominate.
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Explanation: Yes. In an adiabatic process, work transfer can change internal energy.
Answer: Yes.
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Explanation: Yes. At constant volume, heat can change internal energy without boundary work.
Answer: Yes.
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Explanation: Signs and energy transfers depend on what crosses the chosen system boundary.
Answer: System choice defines Q, W and ΔU.
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Explanation: ΔU = 300 - 500 = -200 J, so internal energy decreases.
Answer: ΔU = -200 J.
Quick Revision Notes
Important Formulas
- ΔQ = ΔU + ΔW
- ΔU = ΔQ - ΔW
- W = ∫PdV
- At constant pressure: W = PΔV
- At constant volume: W = 0
- For cyclic process: ΔU = 0
- For cyclic process: Q = W
Important Facts
- Heat and work are path functions.
- Internal energy is a state function.
- Work by system is positive during expansion.
- Work by system is negative during compression.
- Heat supplied is positive; heat rejected is negative.
- Area under P-V curve gives work.
- Area enclosed by P-V loop gives net cyclic work.
Most Common Mistakes
- Using Q = W for every process instead of only when ΔU = 0.
- Forgetting litre to cubic metre conversion in PΔV work.
- Using final pressure instead of graph area when pressure varies.
- Calling heat a stored quantity instead of energy transfer.
- Mixing physics and chemistry sign conventions without stating them.
Exam Tips
- Write the convention before solving: W positive for work done by system.
- For a cycle, immediately write ΔU = 0.
- For a rigid container, immediately write W = 0.
- For adiabatic process, immediately write Q = 0.
- For P-V graph questions, calculate area carefully.
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