For an isolated system, ∆U=0, what will be ∆S ?

Question

The group 14 elements A and B have the first ionisation enthalpy values of 708 and 715 kJ mol$^{-1}$ respectively. The above values are lowest among their group members. The nature of their ions A$^{2+}$ and B$^{4+}$ respectively is:

Answer 1

For an isolated system:

No exchange of heat or work with surroundings: $q = 0,\ w = 0$.
Hence, $\Delta U = 0$ always (energy is conserved).

However, entropy is different:

The second law of thermodynamics states that for an isolated system:
- $\Delta S \ge 0$
- $\Delta S = 0$ for a reversible process (true equilibrium change).
- $\Delta S > 0$ for a spontaneous/irreversible process.

Conclusion:

Even though $\Delta U = 0$ for an isolated system, $\Delta S$ is:

$\Delta S = 0$ if the change is reversible.
$\Delta S > 0$ if the process is spontaneous/irreversible.

So, in general for an isolated system: $\boxed{\Delta S \ge 0}$.

For an isolated system, ∆U=0, what will be ∆S ?

Solution and Explanation

Top Questions on Enthalpy change

Questions Asked in CBSE Class XI exam