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  • One mole of an ideal gas went through an isobaric process then its temperature raised from 27 C to 327 C its entropy change is Option: 1 <img al

One mole of an ideal gas went through an isobaric process then its temperature raised from 27 C to 327 C its entropy change is 

Option: 1

Cv log _e 2


Option: 2

C_p log_e 2


Option: 3

C_p log_e \left ( \frac{1}{2} \right )


Option: 4

C_v log _e \left ( \frac{1}{2} \right )


Answers (1)

best_answer

As we have learned

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

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

 In terms of T and P 

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

For isobaric process P1 = P2 

\Delta S = \mu C_p log _e \left ( \frac{T_2}{T_1} \right )

\Delta S = C_p log _e \left ( \frac{273+327}{273+27} \right )

\Delta S = C_p log _e \left ( 2 \right )

 

Posted by

Nehul

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