Question

Two moles of an ideal gas undergo a reversible isothermal expansion from 2.303l to 23.03l at 27 degree celsius . the change in entropy i

Answer 1

Answer:

Two moles of an ideal gas undergo a reversible isothermal expansion from 2.303l to 23.03l at 27 degree celsius........

Explanation:

..

Answer 2

The entropy change of the process is 38.29 J/K

Explanation:

To calculate the entropy change for isothermal, reversible expansion process, we use the equation:

$\Delta S=nR\ln(\frac{V_2}{V_1})$

where,

$\Delta S$ = Entropy change

n = number of moles = 2 mole

R = Gas constant = 8.314 J/mol.K

$V_1$ = initial volume = 2.303 L

$V_2$ = final volume = 23.03 L

Putting values in above equation, we get:

$\Delta S=2mol\times 8.314J/mol.K\times \ln(\frac{23.03}{2.303})\\\\\Delta S=38.29J/K$

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