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Thermodynamics - First Law of Thermodynamics

Grade 11ICSEPhysics

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

🔑Concepts

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A thermodynamic system is a collection of an extremely large number of atoms or molecules confined within certain boundaries such that it has a pressure PP, volume VV, and temperature TT.

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Internal Energy (UU) is the total energy possessed by the system due to the molecular motion (Kinetic Energy) and molecular configuration (Potential Energy). It is a state function and depends only on the initial and final states.

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First Law of Thermodynamics: It is the law of conservation of energy applied to a thermodynamic system. If an amount of heat ΔQ\Delta Q is supplied to a system, it is used to increase the internal energy ΔU\Delta U and to perform external work ΔW\Delta W.

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Sign Convention (Physics): Heat supplied to the system is +ΔQ+\Delta Q; heat rejected is −ΔQ-\Delta Q. Work done by the system (expansion) is +ΔW+\Delta W; work done on the system (compression) is −ΔW-\Delta W. Increase in internal energy is +ΔU+\Delta U; decrease is −ΔU-\Delta U.

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Molar Specific Heat: The amount of heat required to raise the temperature of 11 mole of a gas by 11 KK (or 1∘C1^\circ C). There are two types: Specific heat at constant volume (CvC_v) and Specific heat at constant pressure (CpC_p).

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Thermodynamic Processes: Isothermal (constant TT, ΔU=0\Delta U = 0), Adiabatic (no heat exchange, ΔQ=0\Delta Q = 0), Isobaric (constant PP), and Isochoric (constant VV, ΔW=0\Delta W = 0).

📐Formulae

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

ΔW=PΔV=∫V1V2P dV\Delta W = P \Delta V = \int_{V_1}^{V_2} P \, dV

ΔU=nCvΔT\Delta U = n C_v \Delta T

Cp−Cv=R (Mayer’s Relation)C_p - C_v = R \text{ (Mayer's Relation)}

γ=CpCv\gamma = \frac{C_p}{C_v}

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

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

💡Examples

Problem 1:

Calculate the change in internal energy of a system when 500 J500 \text{ J} of heat is added to it and the system does 200 J200 \text{ J} of work on the surroundings.

Solution:

Given: ΔQ=+500 J\Delta Q = +500 \text{ J} (heat added), ΔW=+200 J\Delta W = +200 \text{ J} (work done by system). Using First Law: ΔQ=ΔU+ΔW  ⟹  500=ΔU+200  ⟹  ΔU=500−200=300 J\Delta Q = \Delta U + \Delta W \implies 500 = \Delta U + 200 \implies \Delta U = 500 - 200 = 300 \text{ J}.

Explanation:

Since heat is added, ΔQ\Delta Q is positive. Since work is done by the system, ΔW\Delta W is positive. The remaining energy goes into increasing the internal energy of the molecules.

Problem 2:

A gas is compressed at constant pressure of 50 N/m250 \text{ N/m}^2 from a volume of 10 m310 \text{ m}^3 to 4 m34 \text{ m}^3. During this process, 100 J100 \text{ J} of heat is released. Find the change in internal energy.

Solution:

Given: P=50 N/m2P = 50 \text{ N/m}^2, V1=10 m3V_1 = 10 \text{ m}^3, V2=4 m3V_2 = 4 \text{ m}^3, ΔQ=−100 J\Delta Q = -100 \text{ J} (heat released). Work done ΔW=P(V2−V1)=50(4−10)=−300 J\Delta W = P(V_2 - V_1) = 50(4 - 10) = -300 \text{ J}. Using ΔQ=ΔU+ΔW\Delta Q = \Delta U + \Delta W: −100=ΔU+(−300)  ⟹  ΔU=−100+300=200 J-100 = \Delta U + (-300) \implies \Delta U = -100 + 300 = 200 \text{ J}.

Explanation:

Work is negative because the gas is compressed (work done on the system). ΔQ\Delta Q is negative as heat is released. The internal energy increases by 200 J200 \text{ J} despite the heat loss because of the work done on the system.