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First law; work, heat, internal energy and enthalpy: NEET MCQs with answers

75 questions on this topic of Chemical Thermodynamics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following is a path function?

  1. Enthalpy (H)
  2. Internal energy (U)
  3. Heat (q)
  4. Heat capacity (C)
Show answer and explanation

Answer: (C) Heat (q)

Heat (q) and work (w) depend on the path taken to effect the change and are path functions. Enthalpy and internal energy are state functions.

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2. A sample of 2.0 moles of a monoatomic ideal gas (C_v,m = 1.5 R) is heated at constant volume from 300 K to 400 K. The change in enthalpy (ΔH) of the gas is: (Take R = 8.314 J K⁻¹ mol⁻¹)

  1. 2494.2 J
  2. 3325.6 J
  3. 1662.8 J
  4. 4157.0 J
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Answer: (D) 4157.0 J

For a monoatomic ideal gas, C_p,m = C_v,m + R = 1.5 R + R = 2.5 R. The enthalpy change is ΔH = n C_p,m ΔT = 2.0 × (2.5 × 8.314) × (400 - 300) = 5.0 × 8.314 × 100 = 4157.0 J.

hard

3. For the vaporization of 1.0 mol of water at 100 °C and 1.0 bar, the enthalpy of vaporization is 40.66 kJ mol⁻¹. Assuming water vapor behaves as an ideal gas and neglecting the molar volume of liquid water, what is the internal energy change (ΔU) for this process? (Take R = 8.314 J K⁻¹ mol⁻¹)

  1. +40.66 kJ mol⁻¹
  2. +37.56 kJ mol⁻¹
  3. +43.76 kJ mol⁻¹
  4. -37.56 kJ mol⁻¹
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Answer: (B) +37.56 kJ mol⁻¹

For the process H₂O(l) → H₂O(g), Δn_g = 1 - 0 = 1 mol. The relation is ΔH = ΔU + Δn_g RT, so ΔU = ΔH - Δn_g RT. Here, ΔH = 40.66 kJ mol⁻¹ and T = 373.15 K. Δn_g RT = 1 × 8.314 × 373.15 J = 3102.4 J ≈ 3.10 kJ. Thus, ΔU = 40.66 - 3.10 = +37.56 kJ mol⁻¹.

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4. Given below are two statements: Statement I: Heat capacity is an extensive property, whereas specific heat capacity is an intensive property. Statement II: The molar heat capacity of any substance depends on the size or quantity of matter in the sample. In the light of the above statements, choose the most appropriate answer from the options given below:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (B) Statement I is correct but Statement II is incorrect

Heat capacity (C) depends on the mass of the substance and is thus an extensive property. Specific heat capacity (c) and molar heat capacity (C_m) are intensive properties because they are defined per unit mass and per mole, respectively, and are independent of sample size. Hence, Statement I is correct and Statement II is incorrect.

easy

5. Which of the following sets of thermodynamic parameters consists exclusively of extensive properties?

  1. Molar volume, temperature, pressure
  2. Surface tension, viscosity, internal energy
  3. Density, specific heat capacity, enthalpy
  4. Enthalpy, heat capacity, internal energy
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Answer: (D) Enthalpy, heat capacity, internal energy

Extensive properties depend on the quantity of matter present in the system. Enthalpy (H), heat capacity (C), and internal energy (U) are all extensive properties, whereas molar volume, density, temperature, pressure, and specific heat capacity are intensive properties.

hard

6. During an isobaric heating of 1.0 mole of an ideal gas, the heat supplied to the gas is Q. If the ratio of molar heat capacities is γ = C_p / C_v = 1.4, what fraction of the supplied heat is converted into work done by the gas?

  1. 0.29
  2. 0.71
  3. 0.40
  4. 0.50
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Answer: (A) 0.29

At constant pressure, Q = n C_p ΔT = ΔH. The work done by the gas is W = P ΔV = n R ΔT. The fraction of heat converted into expansion work is W / Q = (n R ΔT) / (n C_p ΔT) = R / C_p. Since C_p - C_v = R and γ = C_p / C_v, we have R = C_p (1 - 1/γ) = C_p (γ - 1)/γ. Therefore, W / Q = (γ - 1) / γ = (1.4 - 1) / 1.4 = 0.4 / 1.4 = 2/7 ≈ 0.286 (or 0.29).

medium

7. A gas expands against a constant external pressure of 2 atm by 7 L. Taking 1 L atm = 101.3 J, the work done on the gas is:

  1. −14182 J
  2. 1420 J
  3. −14 J
  4. −1420 J
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Answer: (D) −1420 J

w = −P_ext ΔV = −2 × 7 L atm = −1420 J (negative: the gas does work).

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8. A gas expands against a constant external pressure of 3 atm by 7 L. Taking 1 L atm = 101.3 J, the work done on the gas is:

  1. −21273 J
  2. 2130 J
  3. −21 J
  4. −2130 J
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Answer: (D) −2130 J

w = −P_ext ΔV = −3 × 7 L atm = −2130 J (negative: the gas does work).

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9. A gas expands against a constant external pressure of 2 atm by 4 L. Taking 1 L atm = 101.3 J, the work done on the gas is:

  1. −8104 J
  2. 810 J
  3. −8 J
  4. −810 J
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Answer: (D) −810 J

w = −P_ext ΔV = −2 × 4 L atm = −810 J (negative: the gas does work).

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10. A gas expands against a constant external pressure of 4 atm by 9 L. Taking 1 L atm = 101.3 J, the work done on the gas is:

  1. −3650 J
  2. −36 J
  3. 3650 J
  4. −36468 J
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Answer: (A) −3650 J

w = −P_ext ΔV = −4 × 9 L atm = −3650 J (negative: the gas does work).

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11. For the complete combustion of 1.0 mol of benzene liquid, C₆H₆(l) + 15/2 O₂(g) → 6 CO₂(g) + 3 H₂O(l) at 298 K, what is the value of (ΔH - ΔU)? (Take R = 8.314 J K⁻¹ mol⁻¹)

  1. +1.24 kJ
  2. +3.71 kJ
  3. -1.24 kJ
  4. -3.71 kJ
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Answer: (D) -3.71 kJ

The relation between enthalpy change and internal energy change is ΔH - ΔU = Δn_g RT. For this reaction, Δn_g = moles of gaseous products - moles of gaseous reactants = 6 - 15/2 = -1.5 mol. Thus, ΔH - ΔU = (-1.5 mol) × (8.314 J K⁻¹ mol⁻¹) × (298 K) ≈ -3716 J = -3.71 kJ.

easy

12. Which of the following conditions correctly describes an isolated system?

  1. Exchange of energy with surroundings is allowed, but no exchange of matter
  2. Exchange of both matter and energy with surroundings is allowed
  3. Exchange of matter is allowed, but no exchange of energy
  4. Neither exchange of matter nor exchange of energy with surroundings is allowed
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Answer: (D) Neither exchange of matter nor exchange of energy with surroundings is allowed

An isolated system can exchange neither energy nor matter with its surroundings. A thermos flask with ideal insulation is a common practical approximation.

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