Question

The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10 $d{{m}^{3}}$ to a volume of 100 $d{{m}^{3}}$ at 27$^{0}C$ is:
(A) 38.3 $\text{J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}$
(B) 35.8 $\text{J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}$
(C) 32.3 $\text{J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}$
(D) 42.3 $\text{J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}$

Answer 1

The given substance is a perfect gas i.e. ideal gas. We need to apply the second law of thermodynamics that gives the relation between entropy and heat supplied to a system. When it is an isothermal process, volume is inversely proportional to pressure. After determining the formula, substitute the values in the formula to arrive at the answer.

Complete step-by-step answer:
The second law of thermodynamics states that total entropy of an isolated system never decreases over time, and remains constant for all reversible reaction processes. Isolated systems spontaneously evolve towards thermodynamic equilibrium in order to have maximum entropy.
The total entropy of a system and its surroundings remains constant in ideal conditions i.e. the system has attained thermodynamic equilibrium or going through a reversible process.
The formula for entropy is :
$\Rightarrow$ $\text{ }\\!\\!\Delta\\!\\!\text{ S=}\int\limits_{\text{ }i}^{\text{ }f}{\dfrac{\text{dq}}{\text{T}}}$
Since the temperature remains constant in an isothermal process, we can take the temperature factor out of the integral.
$\Rightarrow$ $\text{ }\\!\\!\Delta\\!\\!\text{ S=}\dfrac{1}{\text{T}}\int\limits_{\text{ }i}^{\text{ }f}{\text{dq}}$
We know that the heat supplied is equal to work done for a reversible isothermal process.
$\Rightarrow$ $\text{d}{{\text{q}}_{\text{rev}\text{.}}}\text{ = }-\text{d}{{\text{w}}_{\text{rev}\text{.}}}\text{ = nRT (}\dfrac{\text{dV}}{\text{V}}\text{)}$
Substituting the value in the formula for entropy we get,
$\Rightarrow$ $\text{ }\\!\\!\Delta\\!\\!\text{ S=}\dfrac{1}{\text{T}}\int\limits_{\text{ }i}^{\text{ }f}{\text{nRT (}\dfrac{\text{dV}}{\text{V}}\text{)}}$
$\Rightarrow$ $\text{ }\\!\\!\Delta\\!\\!\text{ S=}\dfrac{\text{nRT}}{\text{T}}\int\limits_{\text{ }i}^{\text{ }f}{\text{ (}\dfrac{\text{dV}}{\text{V}}\text{)}}$
$\Rightarrow$ $\text{ }\\!\\!\Delta\\!\\!\text{ S= nR ln(}\dfrac{{{V}_{f}}}{{{V}_{i}}})$
Substituting the values in the formula we obtained above, we get
$\Rightarrow$ $\text{ }\\!\\!\Delta\\!\\!\text{ S= 2}\text{.(8}\text{.314) ln(}\dfrac{100}{10})$ = 38.3 $\text{J mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}$

Therefore, the correct answer is option (A).

Note: Thermodynamics deals with equilibrium states and changes from one equilibrium state to another only. We cannot apply thermodynamics law during absence of thermal equilibrium. We can determine the amount of heat transfer with thermodynamics law but it does not give any idea about rate of heat transfer i.e. how long will it take for the completion of the process.