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Thermodynamics - Reversible and Irreversible Processes

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A Reversible Process is a process that can be retraced in the opposite direction such that both the system and the surroundings return to their original states with no other changes anywhere else in the universe.

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Conditions for Reversibility: (i) The process must be Quasi-static (infinitely slow), (ii) There must be no dissipative forces like friction, viscosity, or electrical resistance, and (iii) Heat exchange must occur across an infinitesimal temperature difference ΔT→0\Delta T \to 0.

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An Irreversible Process is one that cannot be undone exactly. All natural processes are irreversible. Factors causing irreversibility include dissipative effects and spontaneous changes (like free expansion) that take the system through non-equilibrium states.

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A Quasi-static process is an idealized process in which the system changes so slowly that it remains in thermal and mechanical equilibrium with its surroundings at every instant.

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The Second Law of Thermodynamics implies that in any irreversible process, the total entropy of the universe increases (ΔStotal>0\Delta S_{total} > 0), whereas for a reversible process, it remains constant (ΔStotal=0\Delta S_{total} = 0).

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In a P−VP-V diagram, a reversible process is represented by a continuous line because every intermediate state is a well-defined equilibrium state.

📐Formulae

Wrev=nRTln⁡(V2V1)=2.303nRTlog⁡10(V2V1)W_{rev} = nRT \ln\left(\frac{V_2}{V_1}\right) = 2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)

ΔS=∫ifdQrevT\Delta S = \int_{i}^{f} \frac{dQ_{rev}}{T}

ηCarnot=1−TLTH\eta_{Carnot} = 1 - \frac{T_L}{T_H}

∮dQT≤0 (Clausius Inequality)\oint \frac{dQ}{T} \le 0 \text{ (Clausius Inequality)}

💡Examples

Problem 1:

Calculate the work done by 2 moles2\text{ moles} of an ideal gas during a reversible isothermal expansion from a volume of 0.02 m30.02\text{ m}^3 to 0.04 m30.04\text{ m}^3 at a constant temperature of 300 K300\text{ K}. (Take R=8.314 J mol−1 K−1R = 8.314\text{ J mol}^{-1}\text{ K}^{-1} and ln⁡(2)≈0.693\ln(2) \approx 0.693)

Solution:

Given: n=2n = 2, T=300 KT = 300\text{ K}, V1=0.02 m3V_1 = 0.02\text{ m}^3, V2=0.04 m3V_2 = 0.04\text{ m}^3. Using the formula for reversible isothermal work: W=nRTln⁡(V2V1)W = nRT \ln\left(\frac{V_2}{V_1}\right) W=2×8.314×300×ln⁡(0.040.02)W = 2 \times 8.314 \times 300 \times \ln\left(\frac{0.04}{0.02}\right) W=4988.4×ln⁡(2)W = 4988.4 \times \ln(2) W=4988.4×0.693≈3457 JW = 4988.4 \times 0.693 \approx 3457\text{ J}

Explanation:

In a reversible isothermal process, the pressure adjusts continuously to stay in equilibrium. The work done is the integral of PdVP dV, which leads to the logarithmic relation with volume ratios.

Problem 2:

A reversible heat engine operates between a source at 800 K800\text{ K} and a sink at 400 K400\text{ K}. If the engine absorbs 1200 J1200\text{ J} of heat from the source, calculate the work done by the engine and the heat rejected to the sink.

Solution:

For a reversible engine: Q1T1=Q2T2\frac{Q_1}{T_1} = \frac{Q_2}{T_2} Given T1=800 KT_1 = 800\text{ K}, T2=400 KT_2 = 400\text{ K}, and Q1=1200 JQ_1 = 1200\text{ J}. Q2=Q1×T2T1=1200×400800=600 JQ_2 = Q_1 \times \frac{T_2}{T_1} = 1200 \times \frac{400}{800} = 600\text{ J} Work done: W=Q1−Q2W = Q_1 - Q_2 W=1200−600=600 JW = 1200 - 600 = 600\text{ J}

Explanation:

In a reversible (Carnot) cycle, the ratio of heat exchanged is proportional to the absolute temperatures of the reservoirs. The energy not converted to work is rejected as heat to the sink.