Question

The entropy of vaporization of benzene is 85 joule per km per hour in 117 gram of engineer prices at is normal water. The entropy change of surrounding is

Answer 1

Answer:

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:

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}}

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