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  • Calculate the change in entropy when 2 moles of an ideal gas expand isotherm

Calculate the change in entropy\mathrm{(\Delta S)} when 2 moles of an ideal gas expand isothermally and reversibly from a volume of 5 L to 20 L at 300 K.

Option: 1

12.05 J/mol-K


Option: 2

22 J/mol-K


Option: 3

23.05 J/mol-K


Option: 4

29 J/mol-K


Answers (1)

best_answer

The entropy change for an isothermal and reversible expansion of an ideal gas
can be calculated using the formula:

\mathrm{\Delta S=n R \ln \left(\frac{V_f}{V_i}\right)}

where n is the number of moles of the gas, R is the ideal gas constant (8.314 J/mol·
K), \mathrm{V_{f}} is the final volume, and \mathrm{V_{i}} is the initial volume.

Given:

\mathrm{\begin{aligned} n & =2 \text { moles } \\ V_i & =5 \mathrm{~L} \\ V_f & =20 \mathrm{~L} \\ R & =8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \end{aligned}}

Substituting the values:

\mathrm{\Delta S=2 \times 8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \times \ln \left(\frac{20}{5}\right) \approx 23.05 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}}

Therefore, the change in entropy (?S) is approximately 23.05 J/mol · K.
So, option c is correct

 

 

Posted by

Pankaj Sanodiya

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