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Question: One mole of a perfect gas expands isothermally to ten times its original volume. The change in entro...

One mole of a perfect gas expands isothermally to ten times its original volume. The change in entropy is:
A. 1.0R1.0R
B. 2.303R2.303R
C. 10.0R10.0R
D. 100.0R100.0R

Explanation

Solution

A perfect gas is the same as ideal gas. It obeys all the gas laws. When a gas expands its atoms tend to get more dispersed in the system thus increasing the volume of gas. We need to find the relation between the entropy and volume of a gas at isothermal conditions.
Formula used:
ΔS=2.303nRlogVfVi\Delta S = 2.303nR\log \dfrac{{{V_f}}}{{{V_i}}}
Where ΔS\Delta S is the change in entropy, nn are the number of moles of gas, RR is the gas constant, Vf{V_f} is the final volume of the system, Vi{V_i} is the initial volume of the system.

Complete step by step answer:
Isothermal process is the process that occurs at constant temperature.
Entropy is the measure of disorder or randomness of molecules in a system. It is a thermodynamic quantity. It is denoted by SS . It is difficult to calculate the absolute entropy of the system. So we calculate the entropy during the change of state. The change in entropy from initial state to final state of a system is given as ΔS\Delta S . Entropy is a state function since it depends on the initial and final state of the system. The entropy change of a system at isothermal reversible conditions is given by-
ΔS=qrevT\Delta S = \dfrac{{{q_{rev}}}}{T}
Where qrev{q_{rev}} is the heat exchanged during the process and TT is the temperature.
The unit of entropy is JK1J{K^{ - 1}} or calK1cal{K^{ - 1}} .
The entropy change for an isothermal process when volume of system changes is given by-
ΔS=2.303nRlogVfVi\Delta S = 2.303nR\log \dfrac{{{V_f}}}{{{V_i}}}
Where ΔS\Delta S is the change in entropy, nn are the number of moles of gas, RR is the gas constant, Vf{V_f} is the final volume of the system, Vi{V_i} is the initial volume of the system.
In the given problem,
Let the initial volume of the system be Vi{V_i} . The volume increases by ten times the initial volume then,
Vf=10Vi{V_f} = 10{V_i} , where Vf{V_f} is the final volume.
And, n=1n = 1
Substituting the values,
 ΔS=2.303nRlogVfVi ΔS=2.303×1×R×log10ViVi ΔS=2.303Rlog10 ΔS=2.303R  \ \Rightarrow \Delta S = 2.303nR\log \dfrac{{{V_f}}}{{{V_i}}} \\\ \Rightarrow \Delta S = 2.303 \times 1 \times R \times \log \dfrac{{10{V_i}}}{{{V_i}}} \\\ \Rightarrow \Delta S = 2.303R\log 10 \\\ \Rightarrow \Delta S = 2.303R \\\ \
The correct option is B.

Note:
-Entropy is a measure of disorder. When entropy increases the system is becoming more disordered from less disordered.
-The entropy for a cyclic process is zero.
-The entropy change in the equilibrium state is zero.