Thermodynamics — Class 11 Physics
1. What Is Thermodynamics?
Thermodynamics studies relationships between heat, work, temperature and energy at the macroscopic level. It does not require us to track the motion of every individual molecule.
A thermodynamic system is the part of the universe chosen for study. Everything outside it is the surroundings. A system may exchange energy and/or matter with its surroundings depending on how it is defined.
2. Thermodynamic Variables
The state of a simple gas can be described using variables such as pressure P, volume V and temperature T. These are state variables. A process describes a change from one equilibrium state to another.
For an ideal gas, the equation of state is PV=nRT, where n is the number of moles and R is the universal gas constant.
3. Thermal Equilibrium and Zeroth Law
The Zeroth Law of Thermodynamics states that if system A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then A and C are in thermal equilibrium with each other. This provides the logical basis for measuring temperature with a thermometer.
4. Heat, Work and Internal Energy
Heat and work are two ways by which energy can cross the boundary of a system. They are not properties stored in a system in the same way as internal energy.
Internal energy U is a state function representing the microscopic energy of the system. For an ideal gas, internal energy depends only on temperature.
5. Sign Convention for Work
In this resource, W denotes work done by the gas. For a quasistatic expansion, W=∫P dV. Expansion gives positive work by the gas; compression gives negative work by the gas.
On a P–V diagram, the work done by the gas in a quasistatic process is the area under the process curve between the initial and final volumes.
6. First Law of Thermodynamics
The first law is the conservation of energy applied to thermodynamic systems:
ΔQ = ΔU + W
where Q is heat supplied to the system, ΔU is the change in internal energy, and W is work done by the system. Equivalently, ΔU=Q−W.
7. Quasistatic Processes
A quasistatic process proceeds sufficiently slowly that the system passes through a sequence of states close to equilibrium. This allows pressure and other state variables to be defined along the path and makes P–V work calculations meaningful.
8. Isothermal Process
In an isothermal process, temperature remains constant. For an ideal gas, internal energy depends only on temperature, so ΔU=0. From the first law, Q=W.
For reversible isothermal expansion of an ideal gas, W=nRT ln(V₂/V₁).
9. Adiabatic Process
In an adiabatic process, there is no heat exchange: Q=0. Therefore, from the first law, ΔU=−W. For a reversible adiabatic process of an ideal gas, PVγ=constant, where γ=Cp/Cv.
10. Isochoric Process
In an isochoric process, volume is constant, so dV=0 and the work done is W=0. Therefore Q=ΔU.
11. Isobaric Process
In an isobaric process, pressure remains constant. For a quasistatic process, W=P(V₂−V₁). Heat supplied is divided between increasing internal energy and doing work.
12. Heat Capacities of a Gas
Molar heat capacity at constant volume is Cv; at constant pressure it is Cp. For an ideal gas, the Mayer relation is Cp−Cv=R. Since a gas at constant pressure expands while being heated, Cp is greater than Cv.
13. Second Law of Thermodynamics
The second law establishes the direction of natural thermodynamic processes and places limits on converting heat into work. Heat does not spontaneously flow from a colder body to a hotter body without external work.
A heat engine cannot convert all the heat absorbed from a single reservoir into work while rejecting no heat. A refrigerator requires work input to transfer heat from a colder region to a hotter region.
14. Heat Engine
A heat engine absorbs heat QH from a hot reservoir, produces work W, and rejects heat QC to a cold reservoir. Energy conservation gives W=QH−QC.
Its efficiency is η=W/QH=1−QC/QH. Efficiency is always less than 1 for a real heat engine.
15. Refrigerator and Coefficient of Performance
A refrigerator removes heat QC from a cold reservoir using work W and rejects QH to the surroundings. Its coefficient of performance is COP=QC/W. Unlike efficiency, COP can be greater than 1.
16. Carnot Engine
A reversible Carnot engine operates between hot temperature TH and cold temperature TC reservoirs. Its ideal efficiency is ηC=1−TC/TH, with temperatures measured on the Kelvin scale.
Worked Numerical 1 — First Law
Question: A gas receives 500 J of heat and does 200 J of work. Find the change in internal energy.
ΔU=Q−W=500−200=300 J.
Worked Numerical 2 — Compression
Question: 150 J of work is done on a gas while 50 J of heat is lost by the gas. Find ΔU.
Heat supplied Q=−50 J. Work done by gas W=−150 J. Hence ΔU=Q−W=−50−(−150)=100 J.
Worked Numerical 3 — Isobaric Work
Question: A gas expands at constant pressure 2×105 Pa from 0.01 m³ to 0.03 m³. Find the work done.
W=P(V₂−V₁)=2×105(0.03−0.01)=4000 J.
Worked Numerical 4 — Isothermal Expansion
Question: One mole of an ideal gas expands reversibly and isothermally at 300 K from volume V to 2V. Find the work done. Take R=8.314 J mol−1 K−1.
W=nRT ln(V₂/V₁)=1×8.314×300×ln2≈1729 J.
Worked Numerical 5 — Isochoric Heating
Question: An ideal gas is heated at constant volume and its internal energy increases by 600 J. Find the work done and heat supplied.
At constant volume, W=0. From ΔU=Q−W, Q=600 J.
Worked Numerical 6 — Heat Engine Efficiency
Question: A heat engine absorbs 2000 J from the hot reservoir and rejects 1200 J. Find its work output and efficiency.
W=QH−QC=2000−1200=800 J.
η=W/QH=800/2000=0.40=40%.
Worked Numerical 7 — Carnot Efficiency
Question: A Carnot engine operates between 500 K and 300 K. Find its ideal efficiency.
η=1−TC/TH=1−300/500=0.40=40%.
Worked Numerical 8 — Refrigerator COP
Question: A refrigerator removes 900 J of heat from its cold compartment using 300 J of work. Find its COP.
COP=QC/W=900/300=3.
Important Derivation — First Law for an Ideal Gas
For a process, conservation of energy requires that heat supplied to a system either increases its internal energy or leaves the system as work done by the system. Therefore Q=ΔU+W. For an ideal gas in an isothermal process, ΔT=0 and hence ΔU=0, giving Q=W.
Important Derivation — Carnot Efficiency
For a reversible Carnot cycle, the heat exchanged during the isothermal stages leads to the relation QC/QH=TC/TH. Since W=QH−QC, efficiency becomes η=1−TC/TH. This is an ideal upper limit for engines operating between those reservoir temperatures.
Explained MCQs
- Which is a state function?
Answer: Internal energy. Its change depends only on initial and final states, whereas heat and work depend on the path. - In an isochoric process, the work done by the gas is:
Answer: Zero. Since volume does not change, ∫P dV=0. - For an ideal gas undergoing an isothermal process, ΔU is:
Answer: Zero. Internal energy of an ideal gas depends only on temperature. - Which is greater for an ideal gas: Cp or Cv?
Answer: Cp. At constant pressure, part of the supplied heat also performs expansion work. - Can a refrigerator have COP greater than 1?
Answer: Yes. COP compares heat removed from the cold region with work input; it is not an efficiency fraction. - For maximum Carnot efficiency between two reservoirs, what should happen to TH and TC?
Answer: The hot-reservoir temperature should be high relative to the cold-reservoir temperature. The ideal expression is η=1−TC/TH.
Competency Question
A pressure cooker raises the boiling temperature of water by increasing pressure above the water. Explain why this changes cooking conditions using the relationship between pressure and boiling, while distinguishing this idea from the first law of thermodynamics.
HOTS
A gas expands adiabatically and does positive work on its surroundings. With Q=0, use the first law to determine the sign of ΔU and explain what happens qualitatively to the gas temperature for an ideal gas.
Common Mistakes
- Mixing sign conventions for work. This resource uses W as work done by the gas.
- Writing Q=ΔU+W without checking what sign has been assigned to W.
- Using Celsius instead of Kelvin in Carnot efficiency.
- Assuming an isothermal process means no heat is transferred; for a gas doing work isothermally, heat must enter to maintain temperature.
- Confusing heat-engine efficiency with refrigerator COP.
- Assuming every thermodynamic process is quasistatic or reversible.
Quick Revision
- PV=nRT
- W=∫P dV for quasistatic work by the gas
- Q=ΔU+W
- Isothermal ideal gas: ΔU=0
- Adiabatic: Q=0
- Isochoric: W=0
- Isobaric: W=P(V₂−V₁)
- Cp−Cv=R
- η=1−QC/QH
- COP=QC/W
- ηCarnot=1−TC/TH