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Thermodynamics - Heat, Internal Energy and Work

Grade 11CBSEPhysics

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

🔑Concepts

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Thermodynamic System: A collection of matter within a boundary that can exchange energy (heat and work) with its surroundings.

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Internal Energy (UU): The sum of the microscopic kinetic and potential energies of the molecules. For an ideal gas, it depends only on temperature TT.

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Heat (QQ): The energy transferred between a system and its surroundings due to a temperature difference.

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Work Done (WW): In thermodynamics, work is defined as the energy transferred by the system to its surroundings through a macroscopic force. For a gas, W=∫V1V2PdVW = \int_{V_1}^{V_2} P dV.

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First Law of Thermodynamics: It is a statement of the law of conservation of energy. It states that the heat energy supplied to a system (QQ) is equal to the sum of the increase in its internal energy (ΔU\Delta U) and the work done by the system on the surroundings (WW): Q=ΔU+WQ = \Delta U + W.

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Sign Convention: QQ is positive if heat is added to the system, and WW is positive if work is done by the system (expansion).

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Mayer's Formula: For an ideal gas, the difference between molar specific heat at constant pressure (CpC_p) and constant volume (CvC_v) is the universal gas constant (RR): Cp−Cv=RC_p - C_v = R.

📐Formulae

Q=ΔU+WQ = \Delta U + W

W=∫V1V2PdVW = \int_{V_1}^{V_2} 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_{\text{isothermal}} = nRT \ln\left(\frac{V_2}{V_1}\right)

Wadiabatic=P1V1−P2V2γ−1W_{\text{adiabatic}} = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}

💡Examples

Problem 1:

A system is provided with 500 J500\text{ J} of heat and it performs 150 J150\text{ J} of work on the surroundings. Calculate the change in the internal energy of the system.

Solution:

Given Q=500 JQ = 500\text{ J} and W=150 JW = 150\text{ J}. According to the First Law of Thermodynamics: Q=ΔU+WQ = \Delta U + W. Rearranging for change in internal energy: ΔU=Q−W\Delta U = Q - W. Substituting the values: 500−150350\begin{array}{r} 500 \\ - 150 \\ \hline 350 \end{array} Therefore, ΔU=350 J\Delta U = 350\text{ J}.

Explanation:

Since heat is added to the system, QQ is positive. Since work is done by the system, WW is positive. The energy remaining after work is performed increases the internal energy.

Problem 2:

Calculate the work done when 22 moles of an ideal gas expand isothermally from a volume of 1 m31\text{ m}^3 to 2 m32\text{ m}^3 at a temperature of 300 K300\text{ K}. (Use R=8.314 J/mol KR = 8.314\text{ J/mol K} and ln⁡(2)≈0.693\ln(2) \approx 0.693)

Solution:

For an isothermal process, the work done is given by W=nRTln⁡(V2V1)W = nRT \ln\left(\frac{V_2}{V_1}\right). Substituting the values: W=2×8.314×300×ln⁡(21)=4988.4×0.693W = 2 \times 8.314 \times 300 \times \ln\left(\frac{2}{1}\right) = 4988.4 \times 0.693. W≈3456.96 JW \approx 3456.96\text{ J}.

Explanation:

In an isothermal expansion, the temperature remains constant, so the internal energy change ΔU=0\Delta U = 0. All the heat supplied to the system is converted into work.