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Thermodynamics - First Law of Thermodynamics

Grade 11CBSEPhysics

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

🔑Concepts

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Internal Energy (UU): The sum of the molecular kinetic and potential energies of a system. For an ideal gas, it depends only on temperature TT. It is a state function, meaning ΔU\Delta U depends only on initial and final states.

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First Law of Thermodynamics: A statement of the law of conservation of energy. It states that the heat supplied to a system (ΔQ\Delta Q) is equal to the sum of the increase in its internal energy (ΔU\Delta U) and the work done by the system (ΔW\Delta W).

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Sign Convention (CBSE/Physics): ΔQ\Delta Q is positive when heat is supplied to the system; ΔW\Delta W is positive when work is done by the system (expansion); ΔU\Delta U is positive when the internal energy/temperature increases.

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Work Done: For a gas undergoing a change in volume from V1V_1 to V2V_2 at pressure PP, the work done is W=∫V1V2PdVW = \int_{V_1}^{V_2} P dV. On a P−VP-V diagram, work done is the area under the curve.

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Mayer's Relation: The relationship between molar specific heat capacity at constant pressure (CpC_p) and constant volume (CvC_v) is given by Cp−Cv=RC_p - C_v = R, where RR is the universal gas constant.

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Specific Heat Capacities: For an ideal gas, the change in internal energy is always ΔU=nCvΔT\Delta U = n C_v \Delta T, regardless of the process.

📐Formulae

ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W

W=∫PdVW = \int P dV

ΔU=nCvΔT\Delta U = n C_v \Delta T

Cp−Cv=RC_p - C_v = R

γ=CpCv\gamma = \frac{C_p}{C_v}

Wisothermal=nRTln⁡(V2V1)W_{isothermal} = nRT \ln\left(\frac{V_2}{V_1}\right)

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

💡Examples

Problem 1:

A system is provided with 200 J200 \text{ J} of heat and the work done by the system is 80 J80 \text{ J}. Calculate the change in internal energy of the system.

Solution:

Given: ΔQ=+200 J\Delta Q = +200 \text{ J} (heat added), ΔW=+80 J\Delta W = +80 \text{ J} (work done by the system). Using the First Law: ΔQ=ΔU+ΔW⇒200=ΔU+80⇒ΔU=200−80=120 J\Delta Q = \Delta U + \Delta W \Rightarrow 200 = \Delta U + 80 \Rightarrow \Delta U = 200 - 80 = 120 \text{ J}.

Explanation:

According to the First Law of Thermodynamics, the energy supplied is partitioned into internal energy increase and mechanical work. Since both heat is added and work is done by the system, the internal energy increases by the difference.

Problem 2:

Find the work done when 2 moles2 \text{ moles} of an ideal gas expand isothermally at 300 K300 \text{ K} from a volume of 10 L10 \text{ L} to 20 L20 \text{ L}. (Use 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:

For an isothermal process, W=nRTln⁡(V2V1)W = nRT \ln\left(\frac{V_2}{V_1}\right). Substituting the values: W=2×8.314×300×ln⁡(2010)=2×8.314×300×0.693≈3457 JW = 2 \times 8.314 \times 300 \times \ln\left(\frac{20}{10}\right) = 2 \times 8.314 \times 300 \times 0.693 \approx 3457 \text{ J}.

Explanation:

In an isothermal process, the temperature remains constant, so ΔU=0\Delta U = 0. Therefore, all the heat absorbed is converted into work done by the gas.

First Law of Thermodynamics - Revision Notes & Key Formulas | CBSE Class 11 Physics