Question

For an irreversible isothermal process occured in a system with temperature T, which following expression best evaluates the change of entropy of the isolated system? (A) ∆S=0 (B) ∆S<0 (C) ∆S>0 (D) ∆S=Q/T (E) ∆S>Q/T (F) ∆S

Answer 1

Answer:

∆S=Q/T

Explanation:

Given the general expression for entropy:

$\Delta S = \int \frac{dQ}{T}$

We can proceed to find the final expression.

As the temperature does not change, it can be removed from the integral, leaving:

$\Delta S=\frac{1}{T} \int dQ$

Now, we proceed to solve the integral to have the final answer:

$\Delta S=\frac{1}{T} Q=\frac{Q}{T} \\\\Therfore\\\\\Delta S=\frac{Q}{T}$