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Thermodynamics - Thermodynamic Processes

Grade 11CBSEPhysics

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

🔑Concepts

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

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An Isothermal Process is a thermodynamic process in which the temperature of the system remains constant (T=constantT = \text{constant}). For an ideal gas, the change in internal energy is zero (ΔU=0\Delta U = 0), and the process follows Boyle's Law: PV=constantPV = \text{constant}.

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An Adiabatic Process is one in which there is no exchange of heat between the system and the surroundings (ΔQ=0\Delta Q = 0). This usually happens when the system is thermally insulated or the process is very fast. It follows the relation PVγ=constantPV^\gamma = \text{constant}.

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An Isobaric Process occurs at constant pressure (P=constantP = \text{constant}). The work done is W=P(V2−V1)W = P(V_2 - V_1) and the heat supplied is used to change both internal energy and perform work.

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An Isochoric Process occurs at constant volume (V=constantV = \text{constant}). Since there is no change in volume, the work done W=0W = 0. By the first law, all heat supplied goes into increasing the internal energy: ΔQ=ΔU\Delta Q = \Delta U.

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The First Law of Thermodynamics is a statement of conservation of energy: ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W, where ΔQ\Delta Q is heat supplied, ΔU\Delta U is change in internal energy, and ΔW\Delta W is work done by the system.

📐Formulae

ΔQ=ΔU+PΔV\Delta Q = \Delta U + P\Delta V

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

PVγ=constant(Adiabatic Process)PV^\gamma = \text{constant} \quad \text{(Adiabatic Process)}

Wadiabatic=nR(T1−T2)γ−1=P1V1−P2V2γ−1W_{\text{adiabatic}} = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{P_1V_1 - P_2V_2}{\gamma - 1}

TVγ−1=constantandP1−γTγ=constantTV^{\gamma-1} = \text{constant} \quad \text{and} \quad P^{1-\gamma}T^\gamma = \text{constant}

γ=CpCv\gamma = \frac{C_p}{C_v}

💡Examples

Problem 1:

Calculate the work done when 1 mole1 \text{ mole} of a gas expands isothermally at 27∘C27^\circ\text{C} to double its original volume. (Given R=8.314 J mol−1K−1R = 8.314 \text{ J mol}^{-1}\text{K}^{-1} and log⁡e2=0.693\log_e 2 = 0.693)

Solution:

Given: n=1n = 1, T=27+273=300 KT = 27 + 273 = 300 \text{ K}, V2=2V1V_2 = 2V_1. Using the isothermal work formula: W=nRTln⁡(V2V1)W = nRT \ln\left(\frac{V_2}{V_1}\right) W=1×8.314×300×ln⁡(2)W = 1 \times 8.314 \times 300 \times \ln(2) W=2494.2×0.693W = 2494.2 \times 0.693 W≈1728.5 JW \approx 1728.5 \text{ J}

Explanation:

Since the process is isothermal, we use the natural log formula for work. Temperature must be converted to Kelvin.

Problem 2:

A gas (γ=1.5)(\gamma = 1.5) at initial pressure PP is compressed suddenly to 14\frac{1}{4}th of its initial volume. Find the final pressure.

Solution:

For a sudden compression, the process is adiabatic. We use: P1V1γ=P2V2γP_1V_1^\gamma = P_2V_2^\gamma P(V)γ=P2(V4)γP(V)^\gamma = P_2\left(\frac{V}{4}\right)^\gamma P2=P×(VV/4)1.5P_2 = P \times \left(\frac{V}{V/4}\right)^{1.5} P2=P×(4)3/2P_2 = P \times (4)^{3/2} P2=P×(4)3=P×8P_2 = P \times (\sqrt{4})^3 = P \times 8 P2=8PP_2 = 8P

Explanation:

The 'sudden' nature implies an adiabatic process. We apply the Poisson's relation PVγ=constantPV^\gamma = \text{constant} to find the ratio of pressures.