When an ideal gas expands into a vacuum:
AInternal energy decreases
BHeat absorbed is maximum
CWork done is zero because $p_\text{ext} = 0$
DWork done is maximum
Answer & Solution
Correct answer: C. Work done is zero because $p_\text{ext} = 0$
$W = -p_\text{ext}\Delta V$. In a vacuum $p_\text{ext} = 0$, so $W = 0$ regardless of the volume change. For an ideal gas this also gives $q = 0$ (since $\Delta U = 0$ at constant temperature for an ideal gas), so the process is called *free expansion*. Both $q$ and $W$ being zero make $\Delta U = 0$ consistent with the isothermal condition.
Related questions
A reaction has $\Delta H = -100$ kJ and $\Delta S = +200$ J/K. At $T = 298$ K, $\Delta G$ Among solid ice, liquid water, and water vapour at the same temperature, entropy is:Hess's law states that the standard enthalpy change for a reaction:For a reaction occurring at constant pressure in an open beaker, the heat absorbed equals:There are two spherical balls $A$ and $B$ of the same material with same surface, but the A cylindrical rod with one end in a steam chamber and the other end in ice results in meltAn ideal gas expands isothermally from a volume $V_{1}$ to $V_{2}$ and then compressed to Two cylinders $A$ and $B$ fitted with pistons contain equal amounts of an ideal diatomic g