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

Hint: To solve this question, we have to remember that entropy is the measure of degree of randomness or disorder in an isolated system and is represented by the symbol S. For an isothermal process, the entropy change is given by, $\vartriangle S = nR\ln \dfrac{{{V_2}}}{{{V_1}}}$

Complete step by step answer:
The greater the degree of randomness, higher is the entropy. In case of chemical reactions, it shows the rearrangement of atoms or ions from one pattern in the reactants to another.
Given that,
n = 1 mole.
Original volume, ${V_1}$ = V.
Final volume, ${V_2}$ = 10V.
We know that, for isothermal process,
Change in entropy, $\vartriangle S = nR\ln \dfrac{{{V_2}}}{{{V_1}}}$ ……. (i)
Putting the given values in equation (i), we will get
$ \Rightarrow \vartriangle S = 1 \times R\ln \dfrac{{10V}}{V}$
As we know that, ln = 2.303 log,
So,
$ \Rightarrow \vartriangle S = 1 \times R \times 2.303{\log _e}10$
Putting ${\log _e}10 = 1$, we will get
 $ \Rightarrow \vartriangle S = 2.303R$
Hence, we can say that the change in entropy will be 2.303R.
So, the correct answer is “Option B”.

Note: If a similar amount of heat is given to a system at lower temperature, and a system at higher temperature, the randomness is more in the system having lower temperature than the system having higher temperature. It shows that entropy change varies inversely with temperature.