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CG PET Thermodynamics — practice questions

Explore 36 free Thermodynamics questions in CG PET Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Thermodynamics questions and exercises with answers

Showing questions 1–30 of 36.

The first law of thermodynamics is mathematically expressed as:Read the answer and worked solution →In an isothermal process for an ideal gas, which thermodynamic quantity remains constant?Read the answer and worked solution →For an ideal gas expanding isothermally from volume $V_1$ to $V_2$ at temperature $T$, the work done by the gas is:Read the answer and worked solution →For an adiabatic process involving an ideal gas, which quantity is constant?Read the answer and worked solution →The first law of thermodynamics is essentially a statement of:Read the answer and worked solution →Work done in an isothermal process for ideal gas:Read the answer and worked solution →In an adiabatic process, the heat exchanged with surroundings is:Read the answer and worked solution →The ratio Cp/Cv is denoted by:Read the answer and worked solution →For an ideal gas, Cp − Cv equals:Read the answer and worked solution →In an ISOCHORIC process, work done equals:Read the answer and worked solution →For an adiabatic process on ideal gas: PV^γ = ?Read the answer and worked solution →Internal energy change ΔU for an ideal gas depends ONLY on:Read the answer and worked solution →In an isothermal expansion of ideal gas, the work done BY the gas is +500 J. The heat absorbed by the gas is:Read the answer and worked solution →A monatomic ideal gas has γ = 5/3. For an adiabatic process, P₁V₁^γ = P₂V₂^γ. If V₂ = 8V₁, find P₂/P₁:Read the answer and worked solution →1 mole of ideal gas (Cv = 3R/2) is heated at constant volume from 300 K to 600 K. Heat absorbed:Read the answer and worked solution →For a monatomic ideal gas in isobaric expansion from V₁ to 3V₁, find ΔT/T₁:Read the answer and worked solution →In a cyclic process, ΔU equals:Read the answer and worked solution →The slope of an isothermal curve in PV diagram (dP/dV)_T equals:Read the answer and worked solution →The slope of an adiabatic curve in PV diagram (dP/dV)_adiabatic equals:Read the answer and worked solution →For a Carnot engine operating between T_h = 600 K and T_c = 300 K, efficiency is:Read the answer and worked solution →A gas absorbs 500 J of heat and does 200 J of work. Change in internal energy:Read the answer and worked solution →For a diatomic ideal gas (γ = 7/5), molar Cv equals:Read the answer and worked solution →Isothermal bulk modulus of an ideal gas equals:Read the answer and worked solution →Adiabatic bulk modulus of ideal gas equals:Read the answer and worked solution →In a free expansion of ideal gas (into vacuum), the temperature:Read the answer and worked solution →An ideal gas undergoes isobaric expansion. Q = 100 J, W = 40 J. Find ΔU:Read the answer and worked solution →For an adiabatic process of an ideal gas, TV^(γ-1) = constant. If T₁ = 300 K and V₂ = 2V₁ for γ = 5/3, then T₂:Read the answer and worked solution →In a Carnot cycle, heat absorbed at hot reservoir is 1000 J. Efficiency 25%. Heat rejected at cold reservoir:Read the answer and worked solution →For polytropic process PV^n = constant, n = 0 means:Read the answer and worked solution →In which process is heat capacity infinite for ideal gas?Read the answer and worked solution →

Worked practice questions

The first law of thermodynamics is mathematically expressed as:

  1. $dQ = dU + dW$
  2. $dQ = dU - dW$
  3. $dU = dQ \cdot dW$
  4. $dQ \cdot dU = dW$
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Correct answer: A. $dQ = dU + dW$

The first law states that heat $dQ$ supplied to a system goes into changing its internal energy $dU$ plus doing work $dW$ on the surroundings. So $dQ = dU + dW$. It is just energy conservation applied to thermodynamic systems.

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In an isothermal process for an ideal gas, which thermodynamic quantity remains constant?

  1. Pressure
  2. Volume
  3. Temperature
  4. Internal energy and temperature
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Correct answer: D. Internal energy and temperature

Isothermal means at constant temperature. For an ideal gas, internal energy depends only on temperature ($U$ is a function of $T$ alone) so if $T$ is constant, $U$ is also constant. Hence both temperature and internal energy stay fixed. Pressure and volume change reciprocally to keep $PV = nRT$ constant.

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For an ideal gas expanding isothermally from volume $V_1$ to $V_2$ at temperature $T$, the work done by the gas is:

  1. $nRT (V_2 - V_1)$
  2. $nRT \ln\dfrac{V_2}{V_1}$
  3. $\dfrac{nRT}{V_2 - V_1}$
  4. $nRT \ln\dfrac{V_1}{V_2}$
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Correct answer: B. $nRT \ln\dfrac{V_2}{V_1}$

For an isothermal process, $P = \dfrac{nRT}{V}$. Work done by the gas equals $\int_{V_1}^{V_2} P\,dV = nRT \int_{V_1}^{V_2} \dfrac{dV}{V} = nRT \ln\dfrac{V_2}{V_1}$. If $V_2 > V_1$ (expansion) the log is positive and the gas does positive work, as it should. Option D flips the limits and gives the wrong sign.

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