Question

Consider a cylinder filled with an ideal gas closed by a frictionless and massless movable piston. The gas undergoes different processes. If $\Delta Q$ represents heat supplied to system, $\Delta W$ work done by gas on surroundings and $\Delta U$ is change in internal energy of the gas then choose the correct statement(s).

• A. For an isothermal expansion and compression $\Delta Q$ is always greater than zero and $\Delta U=0$
• B. For an isothermal expansion and compression $\Delta Q>0$ and $\Delta Q<0$ respectively and $\Delta U=0$ in both cases
• C. In adiabatic expansion $\Delta W>0$ and $\Delta U<0$
• D. In free expansion $\Delta U=\Delta W=\Delta Q=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B For an isothermal expansion and compression $\Delta Q>0$ and $\Delta Q<0$ respectively and $\Delta U=0$ in both cases
C In adiabatic expansion $\Delta W>0$ and $\Delta U<0$
D In free expansion $\Delta U=\Delta W=\Delta Q=0$

In an isothermal process

$\Delta T=0\Rightarrow \Delta U=n{C}_v\Delta T=0$

By first law of thermodynamics,

$\Delta Q=\Delta W$

During expansion $\Delta V>0\Rightarrow \Delta W>0\therefore \Delta Q>0$

During Compression $\Delta V<0\Rightarrow \Delta W<0\therefore \Delta Q<0$

Hence, option B is correct and Option A is wrong.

In an adiabtic process,

$\Delta Q=0\therefore \Delta U=-\Delta W$

since gas is expanding $\Delta W>0$ & $\therefore \Delta U<0$

Hence, Option C is correct.

In free expansion, there will be no external pressure and temperature change.

$\therefore \Delta U=\Delta W=0$

By first law $\Delta Q=0$

Hence, Option D is correct.