Question

One mole of an ideal gas undergoes an isothermal reversible expansion at 25°C. During this process, thesystem absorbs 855 J of heat from the surroundings.When this gas is compressed to the original state in onestep (isothermally), twice as much work is done on thesystem as was performed on the surroundings in theexpansion.
a. What is for the one-step isothermalcompression?
b. What is for the overall process (expansionand compression)?

Short answer

1. $\Delta S$for one-step isothermal compression is$-2.87\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$.
2. $\Delta S_{univ}$for the overall process is$2.871\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$.

Step-by-step solution

Step 1: Introduction to the Concept

The formula calculates the system's change in entropy.

$S=\frac{q_{rev}}{T}$

For a reversible process, $q_{rev}$ is the heat and T is the temperature.

Entropy is a characteristic of a system that quantifies the unpredictability or degree of chaos. It's a state function, after all. With an increase in the universe's entropy, spontaneous change occurs.

$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}$

The entropy changes of the system and its surroundings equal to the entropy change of the universe.

Step 2: Solution Explanation

a)

From the given information,

$q=855\text{J}$

$T=298\text{K}$

So,

$d{S}_{\exp ansion}=\frac{855\text{J}}{298\text{K}}$

$=2.87\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

Let’s determine the change for compression,

$\Delta S_{compression}=-\Delta S_{expansion}$

As a result, the change in compression entropy must be equal to the change in expansion entropy.

So, for one-step isothermal compression, the entropy change is .$-2.87\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

Step 3: Determination of  ΔSuniv

b)

Because entropy is a state function, and the system has been restored to its former state, the change in entropy will be zero.

So, $S_{sys}=0\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

Here, the change in entropy of the surrounding is shown as:

$d{S}_{surr}=\frac{-855\text{J}}{298\text{K}}$

$=-2.7\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

Then,

$S_{univ}=S_{sys}+S_{surr}$

$S=0\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.-2.87\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

$S_{univ}=-2.87\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

Thus, for compression,

$W_{compression}=2(-W_{expansion})$

$=2\left(q_{\exp ansion}\right)$

$=1710\text{J}$

$q_{compression}=-1710\text{J}$

$\Delta S_{compression}=\frac{-1710}{298}$

$=574\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

Let’s compute the total change in entropy of surroundings,

$S_{surr}=-2.87\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.+5.74\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

$=2.871\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$

$\begin{array}{rcl}{S}_{univ}& =& S_{sys}+S_{surr}\\ & =& 0+2.871\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\\ & =& 2.871\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.\end{array}$

Therefore, $\Delta S_{univ}$is $2.871\raisebox{1ex}{$J$}\!\left/ \!\raisebox{-1ex}{$K$}\right.$.