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Indian Navy AA (Artificer Apprentice) Thermodynamics & Kinetic Theory — practice questions

Explore 5 free Thermodynamics & Kinetic Theory questions in Indian Navy AA (Artificer Apprentice) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Thermodynamics & Kinetic Theory questions and exercises with answers

Showing questions 1–5 of 5.

At constant temperature and pressure, the criterion for a process to be **spontaneous** is:Read the answer and worked solution →Hess's law of constant heat summation is a direct consequence of:Read the answer and worked solution →For an adiabatic process, the first law of thermodynamics reduces to:Read the answer and worked solution →System or surroundings?Read the answer and worked solution →Which one does NOT belong?Read the answer and worked solution →

Worked practice questions

At constant temperature and pressure, the criterion for a process to be **spontaneous** is:

  1. $\Delta G < 0$
  2. $\Delta G > 0$
  3. $\Delta S_\text{system} > 0$ only
  4. $\Delta H < 0$ only
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Correct answer: A. $\Delta G < 0$

At constant $T$ and $p$, $\Delta G = \Delta H - T\Delta S$. The second law combined with the first gives that a spontaneous process must have $\Delta G < 0$. $\Delta G = 0$ corresponds to equilibrium and $\Delta G > 0$ is non-spontaneous (the reverse reaction is spontaneous instead).

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Hess's law of constant heat summation is a direct consequence of:

  1. Both the first and second laws together
  2. Energy being a path function
  3. The system being adiabatic
  4. Enthalpy being a state function
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Correct answer: D. Enthalpy being a state function

Hess's law states that the enthalpy change for a reaction depends only on the initial and final states, not on the route taken. This is exactly the statement that enthalpy is a state function. It lets us add and subtract thermochemical equations to obtain the enthalpy of reactions that are difficult to measure directly.

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For an adiabatic process, the first law of thermodynamics reduces to:

  1. $\Delta U = q$, since $W = 0$
  2. $\Delta U = 0$ irrespective of $W$
  3. $\Delta U = W$, since $q = 0$
  4. $q = -W$, so $\Delta U = 0$
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Correct answer: C. $\Delta U = W$, since $q = 0$

In an adiabatic process no heat is exchanged with the surroundings, $q = 0$. The first law $\Delta U = q + W$ therefore becomes $\Delta U = W$. So an adiabatic compression ($W > 0$) increases internal energy and hence temperature an adiabatic expansion ($W < 0$) decreases both.

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