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  • One mole of an ideal gas when went through a thermodynamic process its pressure and volume changed from p and V to 2P and 2V then its entropy change is equal to <st

One mole of an ideal gas when went through a thermodynamic process its pressure and volume changed from p and V to 2P and 2V then its entropy change is equal to 

Option: 1

Cv log _e 2


Option: 2

Cp log _e 2


Option: 3

(Cp+Cv) log_e 2


Option: 4

(Cp-Cv ) log _e 2


Answers (1)

best_answer

As we have learned

Entropy change in terms of Pressure and Volume -

\Delta S= nc_{v}\ln \left ( \frac{P_{2}}{P_{1}} \right )+n\, C_{P}\ln \left ( \frac{V_{2}}{V_{1}} \right )

-

 

 enrtopy change in term of P and V is 

\Delta S= nc_{v}\ln \left ( \frac{P_{2}}{P_{1}} \right )+n\, C_{P}\ln \left ( \frac{V_{2}}{V_{1}} \right )

 

\Delta S= nc_{v}\ln \left ( \frac{2P}{P_{1}} \right )+n\, C_{P}\ln \left ( \frac{2V}{V_{1}} \right )

 

\Delta S= c_{v} log _e (2 )+ C_{P} log _e (2)

\Delta S=log _e (2) \times (C_v + C_P )

Posted by

Deependra Verma

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