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  • The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from initial position to final state is equal to Option: 1</s

The change in the entropy of a 1 mole of an ideal gas which went through an isothermal process from initial position to final state is equal to 

Option: 1

Rln \frac{V_2}{V_1}


Option: 2

Rln \frac{V_1}{V_2}


Option: 3

zero 


Option: 4

R ln T 


Answers (1)

best_answer

As we have learned

Entropy change for an ideal gas in terms of T & V -

\Delta S= n\, c_{p}\ln \left ( \frac{T_{2}}{T_{1}} \right )+nR\ln \left ( \frac{V_{2}}{V_{1}} \right )

 For an ideal gas, we have 

\Delta S= n\, c_{p}\ln \left ( \frac{T_{2}}{T_{1}} \right )+nR\ln \left ( \frac{V_{2}}{V_{1}} \right )

for isothermal process 

log _e \left ( \frac{T_2}{T_1 } \right )= 0

So \Delta s = n R log _e \left ( \frac{V_2}{V_1} \right )   for one mole of gas 

\Delta s =R log _e \left ( \frac{V_2 }{V_1 } \right )

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