Question

Entropy of vaporization of benzene is 85 when 117 gm benzene vapourizes at its normal boiling point calculate entropy change of surrounding

Answer 1

Answer: The entropy of the surrounding is -126.65 J/K.

Explanation: We are given entropy of vaporization of benzene, which is 85 J/K mol.

To calculate the moles, we use the formula:

$Moles=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of benzene = 117 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

$Moles=\frac{117g}{78.11g/mol}=1.49mol$

For 1 mole of benzene, the entropy of vaporization is 85 J/K mol

So, for 1.49 moles of benzene, the entropy of vaporization will be = $1.49\times 85=126.65J/K$

Now, at boiling point, benzene is in equilibrium with its vapor. Therefore, the two phases are in equilibrium and for such processes:

$\Delta S_{system}+\Delta S_{surrounding}=0$

$\Delta S_{surrounding}=-\Delta S_{system}=-126.65J/K$

Answer 2

Answer: The entropy of the surrounding is -126.65 J/K.

Explanation: We are given entropy of vaporization of benzene, which is 85 J/K mol.

To calculate the moles, we use the formula:

Moles=\frac{\text{Given mass}}{\text{Molar mass}}Moles=Molar massGiven mass​

Given mass of benzene = 117 g

Molar mass of benzene = 78.11 g/mol

Putting values in above equation, we get:

Moles=\frac{117g}{78.11g/mol}=1.49molMoles=78.11g/mol117g​=1.49mol

For 1 mole of benzene, the entropy of vaporization is 85 J/K mol

So, for 1.49 moles of benzene, the entropy of vaporization will be = 1.49\times 85=126.65J/K1.49×85=126.65J/K

Now, at boiling point, benzene is in equilibrium with its vapor. Therefore, the two phases are in equilibrium and for such processes:

\Delta S_{system}+\Delta S_{surrounding}=0ΔSsystem​+ΔSsurrounding​=0

\Delta S_{surrounding}=-\Delta S_{system}=-126.65J/KΔSsurrounding​=−ΔSsystem​=−126.65J/K