Question

One mole of perfect gas expands isothermally to ten times its original volume. The change in entropy is:
A. 0.1R
B. 2.303R
C. 10.0R
D. 100.0R

Answer 1

In question, it is given that the process is isothermal and final volume of gas becomes ten times of the original volume. We need to calculate the change in entropy, and for that firstly know what exactly entropy is and how it depends on change in volume and change in temperature. Put values in mathematical expression for entropy for isothermal processes and calculate change in entropy.

Formula Used:
$\Delta S = {{\text{C}}_{\text{v}}}{\text{ln}}\dfrac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ + 2}}{\text{.303Rlog}}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}}$
$\Delta S = 2{\text{.303Rlog}}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}}$

Complete step by step answer:
In thermodynamics, we studied that if a system is perfectly conducting with the surrounding, and undergoes a process in such a way that the temperature remains constant throughout the process is called an isothermal process. For instance, vaporization of water at its boiling point.

In this question, we are given that a perfect gas expands isothermally, therefore, expansion of gas is carried out by keeping temperature constant. But, during expansion cooling is produced but at the same time heat would enter from the walls and for allowing heat to enter we need to keep the walls of the container of gas conducting. Such that, there would not be any change in the temperature of gas.Therefore, if ‘T’ is assumed to be temperature of gas and process is isothermal,
$\Rightarrow \Delta T = 0$(Change in temperature is zero)

Coming back to question, it says that after expansion the volume of gas is equal to ten times of its original volume. So if ‘${{\text{V}}_{\text{1}}}$’ is volume of gas before expansion and ‘${{\text{V}}_{\text{2}}}$’ is the volume of gas after expansion.
Then, according to question,
$\Rightarrow {{\text{V}}_{\text{2}}} = 10{{\text{V}}_{\text{1}}}$
We need to calculate the change in entropy for this process, and for that firstly we should know what exactly entropy is and how entropy depends on change in volume and change in temperature throughout the process.

Entropy in most simple words is known as the degree of randomness or disorder of molecules or particles of matter. Greater the randomness of a system, higher would be the entropy of a thermodynamically system. It is denoted by the symbol ‘S’. Now you can easily understand that, if there is expansion, definitely the molecules would move rapidly and entropy would have changed. And mathematically, it is given by:
$\Rightarrow\Delta S =\int {\dfrac{{{\text{d}}{{\text{Q}}_{{\text{rev}}}}}}{{\text{T}}}}$………..Eq.1
Where, $\Delta S$= change in entropy
${\text{d}}{{\text{Q}}_{{\text{rev}}}}$= change in heat for reversible process
Also, we learnt a derivation in unit-Chemical Thermodynamics, for change in entropy for isothermal process by substituting values of dQ from first law of thermodynamics which is:
$\Rightarrow\Delta S = {{\text{C}}_{\text{v}}}{\text{ln}}\dfrac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ + Rln}}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}}$
$\Rightarrow\Delta S = {{\text{C}}_{\text{v}}}{\text{ln}}\dfrac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ + 2}}{\text{.303Rlog}}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}}$…………Eq.2

Since, $\Delta T = 0$ so this term ${{\text{C}}_{\text{v}}}{\text{ln}}\dfrac{{{{\text{T}}_{\text{2}}}}}{{{{\text{T}}_{\text{1}}}}}$in Eq.2 would become zero and,
$\Rightarrow\Delta S = 2{\text{.303Rlog}}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{V}}_{\text{1}}}}}$…………..Eq.3
Substituting values, ${{\text{V}}_{\text{2}}} = 10{{\text{V}}_{\text{1}}}$ in Eq.3, we get

\Rightarrow\Delta S = 2{\text{.303Rlog}}\dfrac{{{\text{10}}{{\text{V}}_1}}}{{{{\text{V}}_{\text{1}}}}} \\\ \Rightarrow\Delta S = 2{\text{.303Rlog(10)}} \\\ $$ And, $${\text{log(10)}} = 1$$ $$ \therefore\Delta S = 2{\text{.303R}}$$ Therefore, the change in entropy is 2.303R. **Hence, option B is correct.** **Additional Information:** Entropy is a state function and an extensive property. Whenever the state of a system changes, the entropy of the system does undergo some change. Only change in entropy can be calculated, the exact value of entropy for any instant of time can’t be measured. **Note:** ln (natural log) in mathematical equation should be converted into log with base 10. Also the numerical value of ln (1) is zero. Go through definitions of types of processes like isobaric, adiabatic, and isothermal processes properly as it can cause confusion. Learn derivations properly and their physical significance for proper conceptual information.