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

∆S=Q/T

Step-by-step explanation:

Given the general expression for entropy:

`\Delta S = \int (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=(1)/(T) \int dQ`

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

`\Delta S=(1)/(T) Q=(Q)/(T) \\\\Therfore\\\\\Delta S=(Q)/(T)`