Respuesta :
Answer : The change in molar entropy of the sample is -14.22 J/K.mol
Explanation :
To calculate the change in molar entropy we use the formula:
[tex]\Delta S=n\times C_v\times \ln (\frac{T_2}{T_1})+n\times R\times \ln (\frac{V_2}{V_1})[/tex]
where,
[tex]\Delta S[/tex] = change in molar entropy
n = number of moles = 1 mole
[tex]T_2[/tex] = final temperature = [tex]28.1^oC=273+28.1=301.1K[/tex]
[tex]T_1[/tex] = initial temperature = [tex]18.5^oC=273+18.5=291.5K[/tex]
[tex]V_2[/tex] = final volume = 0.500 L
[tex]V_1[/tex] = initial volume = 3.00 L
[tex]C_{v}[/tex] = heat capacity diatomic gas [tex](N_2)[/tex] = [tex]\frac{5}{2}R[/tex]
R = gas constant = 8.314 J/mol.K
Now put all the given values in the above formula, we get:
[tex]\Delta S=1\times (\frac{5}{2}R)\times \ln (\frac{301.1}{291.5})+1\times R\times \ln (\frac{0.500}{3.00})[/tex]
[tex]\Delta S=1\times (\frac{5}{2}\times 8.314)\times \ln (\frac{301.1}{291.5})+1\times 8.314\times \ln (\frac{0.500}{3.00})[/tex]
[tex]\Delta S=-14.22J/K.mol[/tex]
Therefore, the change in molar entropy of the sample is -14.22 J/K.mol
