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  • How to solve this problem- The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10dm3to a volume of 100dm3at 27°C is:

The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10dm3 to a volume of 100dm3 at 27°C is:

  • Option 1)

    38.3 J mol-1 K-1

  • Option 2)

    35.8 J mol-1 K-1

  • Option 3)

    32.3 J mol-1 K-1

  • Option 4)

    42.3 J mol-1 K-1

 

Answers (1)

best_answer

 

Entropy for isothermal process -

\Delta S= nR\ln \frac{V_{f}}{V_{i}}

or

\Delta S= nR\ln \frac{P_{i}}{P_{f}}
 

- wherein

T_{f}=T_{i}

\Delta T=0

 

 \Delta S= nRln \frac{V1}{V2}=2.303 nR \log \frac{V_{1}}{V_{2}}

=2.303 \times 2 \times 8.314 \times log \frac{100}{10}

= 38.3 J mol-1K-1

 


Option 1)

38.3 J mol-1 K-1

correct

Option 2)

35.8 J mol-1 K-1

Incorrect

Option 3)

32.3 J mol-1 K-1

Incorrect

Option 4)

42.3 J mol-1 K-1

Incorrect

Posted by

divya.saini

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