Question

1 mol of diatomic ideal gas expands from 1 L to 10 L
in reversible isothermal process. The entropy
change AS of surrounding for this process​

Answer 1

Answer: The entropy change of the process is 19.14 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 = 1 mole

R = Gas constant = 8.314 J/mol.K

$V_1$ = initial volume = 1 L

$V_2$ = final volume = 10 L

Putting values in above equation, we get:

$\Delta S = 1mol\times 8.314J/mol.K\times \ln(\frac{10}{1})\\\\\Delta S=19.14J/K$

Hence, the entropy change of the process is 19.14 J/K