Question

a liquid is at the boiling point of 289°c . The part of the total mass say 0.9 kg of the Mercury vapour condense to the liquid. Calculate the entropy change of the vapour ​

Answer 1

Answer:

4.1kg

Explanation:

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Answer 2

Answer:

To calculate the entropy change of the vapour, we need to use the formula:

ΔS = q/T

Explanation:

where ΔS is the entropy change, q is the heat transferred, and T is the temperature in Kelvin. We can find q by using the heat of condensation of mercury, which is 58.7 kJ/mol. Since we know the mass of the vapour that condensed (0.9 kg), we can convert this to moles using the molar mass of mercury (200.59 g/mol):

0.9 kg / 200.59 g/mol = 0.00449 mol

Then, we can calculate the heat transferred using the following:

q = ΔHcondensation × n

Where n is the number of moles that condensed:

q = 58.7 kJ/mol × 0.00449 mol = 0.263 kJ

Finally, we can calculate the entropy change using:

ΔS = q/T

We need to convert the boiling point of mercury to Kelvin:

289°C + 273.15 = 562.15 K

ΔS = 0.263 kJ / 562.15 K = 0.000467 kJ/K

Therefore, the entropy change of the vapour is 0.000467 kJ/K. This value represents the increase in disorder of the system as a result of the condensation process. The entropy change is small because the condensation only involved a small portion of the total mass of the vapor.

To learn more about entropy, click on the given link.

https://brainly.in/question/525552

To learn more about vapours, click on the given link.

https://brainly.in/question/49753571

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