Question

For 1 mole of an ideal monatomic gas on moving from one state to another, the temperature is doubled but pressure becomes $\sqrt {\text{2}}$ times. The entropy change in the process will be $\left( {R = 2{\text{ cal/mol}} \cdot {\text{K}}} \right)$
(A) $R{\text{ ln2}}$
(B) ${\text{2}}R{\text{ ln2}}$
(C) ${\text{3}}R{\text{ ln2}}$
(D) $\dfrac{R}{2}{\text{ ln2}}$

Answer 1

The measure of randomness or disordered distribution is known as entropy. The randomness is always higher in a gaseous state. More the number of gaseous molecules higher is the entropy.

Complete step by step solution:
First we will derive an equation for entropy in terms of temperature and pressure.
The equation for the change in entropy is as follows:
$\Delta S = \dfrac{{\Delta Q}}{T}$ ………..…… (1)
Where $\Delta S$ is the change in entropy,
$\Delta Q$ is the change in the heat of the system,
$T$ is the temperature at which the reaction occurs.
The equation for the change in heat is as follows:
$\Delta Q = \Delta H - V\Delta P$ ………..…… (2)
Where, $\Delta Q$ is the change in the heat of the system,
$\Delta H$ is the change in the enthalpy of the system,
$V$ is the volume,
$\Delta P$ is the change in the pressure.
The equation for the change in enthalpy is as follows:
$\Delta H = {C_P}\Delta T$ ……...…… (3)
Where, $\Delta H$ is the change in the enthalpy of the system,
${C_P}$ is the heat capacity at the constant pressure,
$\Delta T$ is the change in the temperature.
From equation (1), equation (2) and equation (3),
$\Delta S = \dfrac{{{C_P}\Delta T - V\Delta P}}{T}$
$\Rightarrow \Delta S = \dfrac{{{C_P}\Delta T}}{T} - \dfrac{{V\Delta P}}{T}$
$\Rightarrow \Delta S = \dfrac{{{C_P}\Delta T}}{T} - \dfrac{{R\Delta P}}{P}$ (From the ideal gas equation, $V{\text{/}}T = R{\text{/}}P$)
$\Delta S = {C_P}\dfrac{{\Delta T}}{T} - R\dfrac{{\Delta P}}{P}$
Thus,
$\Delta S = {C_P}\ln \dfrac{{{T_2}}}{{{T_1}}} - R\ln \dfrac{{{P_2}}}{{{P_1}}}$ …… (4)
We know that for an ideal monatomic gas,
${C_P} = \dfrac{{5R}}{2}$ ………………... (5)
From equation (4) and equation (5),
$\Delta S = \dfrac{{5R}}{2}\ln \dfrac{{{T_2}}}{{{T_1}}} - R\ln \dfrac{{{P_2}}}{{{P_1}}}$ ………….…… (6)
Equation (6) is the equation for entropy in terms of temperature and pressure.
We are given that the temperature is doubled thus, ${T_2} = 2 \times {T_1}$. The pressure becomes $\sqrt {\text{2}}$ times thus, ${P_2} = \sqrt 2 \times {P_1}$. Thus, equation (6) becomes as follows:
$\Delta S = \dfrac{{5R}}{2}\ln \dfrac{{2 \times {T_1}}}{{{T_1}}} - R \times \ln \dfrac{{\sqrt 2 \times {P_1}}}{{{P_1}}}$
$\Delta S = \dfrac{{5R}}{2}\ln 2 - R \times \ln \sqrt 2$
$\Rightarrow \Delta S = \dfrac{{5R}}{2}\ln 2 - R \times \ln {\left( 2 \right)^{{\text{1/2}}}}$
$\Rightarrow \Delta S = \dfrac{{5R}}{2}\ln 2 - \dfrac{R}{2} \times \ln 2$
$\Rightarrow\Delta S = \left( {\dfrac{5}{2} - \dfrac{1}{2}} \right)R\ln 2$
$\Rightarrow \Delta S = \left( {\dfrac{4}{2}} \right)R\ln 2$
$\Rightarrow \Delta S = {\text{2}}R{\text{ ln2}}$
Thus, the entropy change in the process will be ${\text{2}}R{\text{ ln2}}$.

Thus, the correct option is (B) ${\text{2}}R{\text{ ln2}}$.

Note: The heat capacity at constant pressure contributes to the work done as well as the change in internal energy. It is denoted by ${C_P}$. For a monatomic ideal gas, the heat capacity at constant pressure is equal to $\dfrac{{5R}}{2}$ where $R$ is the universal gas constant.