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JKCET Thermodynamics — practice questions

Explore 13 free Thermodynamics questions in JKCET Chemistry. 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 1–13 of 13.

Worked practice questions

Which of the following is an example of an **isolated** thermodynamic system?

  1. Hot tea in an open cup placed on a table
  2. A reaction mixture in a sealed copper vessel
  3. Boiling water in a closed glass beaker
  4. Hot coffee in a thermos flask sealed with its lid
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Correct answer: D. Hot coffee in a thermos flask sealed with its lid

An isolated system permits neither matter nor energy exchange with the surroundings. A well-sealed thermos flask is designed precisely for this — vacuum jacket blocks heat transfer, the lid blocks matter exchange. The open cup is an open system (matter + energy). The sealed copper vessel is a closed system (no matter exchange, but copper conducts heat). The closed glass beaker is also closed (glass still conducts heat).

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The mathematical statement of the **first law of thermodynamics** is:

  1. $\Delta U = q - w$
  2. $\Delta U = q + w$
  3. $\Delta H = q_v$
  4. $\Delta G = \Delta H - T\Delta S$
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Correct answer: B. $\Delta U = q + w$

The first law (with the IUPAC convention that w is work done **on** the system) reads $\Delta U = q + w$, expressing the conservation of energy for an isolated system. $\Delta U = q - w$ uses the older convention where w is work done **by** the system. $\Delta H = q_p$ (constant pressure, not $q_v$) and the Gibbs relation are different state equations.

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For the **isothermal free expansion** of an ideal gas into vacuum, which set of statements is correct?

  1. q > 0, w < 0, ΔU > 0
  2. q = 0, w = 0, ΔU = 0
  3. q < 0, w > 0, ΔU = 0
  4. q = 0, w > 0, ΔU > 0
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Correct answer: B. q = 0, w = 0, ΔU = 0

Free expansion occurs against zero external pressure, so $w = -p_{ex}\Delta V = 0$. Joule's experiment showed that for an ideal gas, expansion into vacuum produces no temperature change, so $q = 0$. From the first law, $\Delta U = q + w = 0$. Internal energy of an ideal gas depends only on T, which is unchanged here, confirming ΔU = 0.

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