Answer : The change in entropy is, 8.65 J/K
Explanation :
Formula used :
[tex]\Delta S=nC_v\ln (\frac{T_2}{T_1})+nR\ln (\frac{V_2}{V_1})[/tex]
where,
[tex]\Delta S[/tex] = change in entropy
n = number of moles of gas = 1 mole
R = gas constant = 8.314 J/mol.K
[tex]C_v[/tex] = specific heat capacity at constant volume = [tex]\frac{5}{2}R[/tex] (for monoatomic gas)
[tex]V_1[/tex] = initial volume of gas = 1.3 L
[tex]V_2[/tex] = final volume of gas = 3.1 L
[tex]T_1[/tex] = initial volume of gas = [tex]23^oC=273+23=296K[/tex]
[tex]T_2[/tex] = final volume of gas = [tex]44^oC=273+44=317K[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta S=1\times \frac{5}{2}R\ln (\frac{317}{296})+1\times R\ln (\frac{3.1}{1.3})[/tex]
[tex]\Delta S=1\times \frac{5}{2}\times 8.314\ln (\frac{317}{296})+1\times 8.314\ln (\frac{3.1}{1.3})[/tex]
[tex]\Delta S=8.65J/K[/tex]
Therefore, the change in entropy is, 8.65 J/K
