- Introduction
- Concept of Heat
- P-V Indicator Digram
- |
- Work in volume changes
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- Internal Energy and first law of thermodynamics
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- Specific heat capacity of an ideal gas
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- Thermodynamic Processes
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- Quasi static Processes
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- Isothermal Process
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- Adiabatic Process
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- Isochoric process
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- Isobaric process
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- Work done in Isothermal process
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- Work done in an Adiabatic process
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- Heat Engine and efficiency
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- Principle of a Refrigerator
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- Second law of thermodynamics
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- Reversibility and irreversibility
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- Carnot's Heat Engine
- |
- Carnot Theorem
- |
- Solved Examples

- In an isothermal process temperature remains constant.

- Consider pressure and volume of ideal gas changes from (P
_{1}, V_{1}) to (P_{2}, V_{2}) then, from first law of thermodynamics

ΔW = PΔV

Now taking ΔV aproaching zero i.e. ΔV and suming ΔW over entire process we get total work done by gas so we have

W = ∫PdV

where limits of integration goes from V_{1}to V_{2}

as PV = nRT we have P = nRT / V

W = ∫(nRT/V)dV

where limits of integration goes from V_{1}to V_{2}

on integrating we get,

W=nRT ln(V_{2}/V_{1}) (3)

Where n is number of moles in sample of gas taken.

- For an adiabatic process of ideal gas equation we have

PV^{γ}= K (Constant) (14)

Where γ is the ratio of specific heat (ordinary or molar) at constant pressure and at constant voluume

γ = C_{p}/C_{v}

- Suppose in an adiabatic process pressure and volume of a sample of gas changs from (P
_{1}, V_{1}) to (P_{2}, V_{2}) then we have

P_{1}(V_{1})^{γ}=P_{2}(V_{2})^{γ}=K

Thus, P = K/V^{γ}

- Work done by gas in this process is

W = ∫PdV

where limits of integration goes from V_{1}to V_{2}

Putting for P=K/V^{γ}, and integrating we get,

W = (P_{1}V_{1}-P_{2}V_{2})/(γ-1) (16)

- In and adiabatic process if W>0 i.e., work is done by the gas then T
_{2}< T_{1}

- If work is done on the gas (W<0) then T
_{2}> T_{1}i.e., temperature of gas rises.

- Any device which convents heat continously into mechenical work is called a heat engine.

- For any heat engine there are three essential requirements.

(i) SOURCE : A hot body at fixed temperature T_{1}from which heat engine can draw heat

(ii) Sink : A cold body, at a fixed lower temprature T_{2}, to which any amount of heat can be rijectd.

(iii) WOEKING SUBTANCE : The material, which on being supplied with heat will do mechanical work.

- In heat engine, working substances, could be gas in cylinder with a moving piston.

- In heat engine working substance takes heat from the sorce, convents a part of it into mechanical work, gives out rest to the sink and returns to the initial state. This series of operations constitutes a cycle.

- This cycle is represented in fig below

- Work from heat engine can be continously obtained by performing same cycle again and again.

- Consider,

Q_{1}- heat absorbed by working substance from sorce

Q_{2}- heat rejected to the since

W - net amount of work done by working substance

Q_{1}-Q_{2}- net amount of heat absorbed by working substance.

ΔU = 0 since in the cycle Working Substance returns to its initial condition.

So on application of first law of thermodynamics

Q_{1}-Q_{2}= W

- Thermal efficiency of heat engine

η= work output in energy units / Heat input in same energy units

= W / Q_{1}= (Q_{1}-Q_{2})/ Q_{1}

Or, η = 1-(Q_{2}/Q_{1}) (17)

from this equation it is clear that

Q = 1 for Q_{2}=0

and there would be 100% conversion of heat absorbed into work but such ideal engines are not possible in practice.

Class 11 Mathse Class 11 Physics

- Principles of Physics Extended (International Student Version) (WSE)
- university physics with modern physics by Hugh D. Young 13th edition
- NCERT Exemplar Problems: Solutions Physics Class 11
- H.C. Verma Concepts of Physics - Vol. 1
- CBSE All in One Physics Class 11 by Arihant
- NCERT Solutions: Physics Class 11th
- New Simplified Physics: A Reference Book for Class 11 (Set of 2 Parts)
- Pradeep's A Text Book of Physics with Value Based Questions - Class XI (Set of 2 Volumes)
- Oswaal CBSE Sample Question Papers For Class 11 Physics (For 2016 Exams)

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