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# The entropy change when 1 mole of ideal gas at 27$^o$C is expanded reversibly from 2 litre to 20 litre is (A) $2cal{K^{ - 1}}mo{l^{ - 1}}$ (B) $4.6cal{K^{ - 1}}mo{l^{ - 1}}$ (C) $4J{K^{ - 1}}mo{l^{ - 1}}$ (D) $4.6kJ{K^{ - 1}}mo{l^{ - 1}}$ Verified
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Hint :Entropy of the system is defined as the randomness or degree of freedom of any system. Entropy is denoted by ‘S’. Entropy is a thermodynamic quantity which depends on thermodynamic variables that are temperature, pressure and volume. We cannot calculate the absolute value of entropy for any system but we can calculate the change in entropy
( $\Delta S$ ).

Thermodynamically change in entropy is represented as
$\Delta S = \dfrac{{{q_{reversible}}}}{T}$ . This is the formula to calculate change in entropy.
Where,
${q_{reversible}}$ = heat exchange for the system which is reversible in nature (Joule or calorie).
$T$ = temperature(K).
The unit of entropy is $J{K^{ - 1}}$ or $Cal{K^{ - 1}}$ .
$1J{K^{ - 1}}$ is said to be 1 entropy unit.
Given: $\Delta S$ =?
n=1mol
T= ${27^0}C$ = $27 + 273 = 300K$ .
${V_1} = 2L$
${V_2} = 20L$
Let’s consider $R = 2calmo{l^{ - 1}}{K^{ - 1}}$
Let’s rewrite the equation,
$\Delta S = \dfrac{{{q_{reversible}}}}{T}$
Substituting q by $\Delta U + P\Delta V$ ,
$\Delta S = \dfrac{{\Delta U + P\Delta V}}{T}$
For an ideal gas $\Delta T = 0,\Delta U = 0$ .
So, we get
$\Delta S = \dfrac{{P\Delta V}}{T}$
From the ideal gas equation we know that $P = \dfrac{{nrT}}{V}$
$\Delta S = \dfrac{{nRT}}{{V \times T}} \times \Delta V$
$\Delta S = nR\dfrac{{\Delta V}}{V}$
On integration we have,
$\Delta S = nR\ln \dfrac{{{V_2}}}{{{V_1}}}$ .
Substituting the given values we have,
$\Rightarrow \Delta S = 1 \times 2 \times \ln \dfrac{{20}}{2}$
$\Delta S = 2 \times 2.303$
$\Delta S = 4.606cal{K^{ - 1}}mo{l^{ - 1}}$ .
So, the correct option is (B).

Note :
For any reversible system there is one and only path. So we can say that ${q_{reversible}}$ is a state function. Change in entropy is inversely proportional to temperature that is at low