Question

the enthalpy of vapourisation of liquid is 30 kJ mol -1 and the entropy of vapourisation is 75 KJ mol -1. The boiling point of the liquud at 1 atm is

a) 250 K
b) 400 K
c) 450 K
d) 600K

*The answer is 400K*

please explain how it came...
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Answer 1

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Answer 2

Concept Introduction: Thermodynamics is a natural and universal process.

Given:

We have been Given: Enthalpy of vapourisation is

$30kj {mol}^{ - 1}$

Entropy of Vapourisation is

$75kj {mol}^{ - 1}$

To Find:

We have to Find: The boiling point of the liquud at 1 atm is:

1. 250 K
2. 400 K
3. 450 K
4. 600K

Solution:

According to the problem,

$s_{vp} = \frac{ h_{vp} }{ t_{b} } \\ t_{b} = \frac{ h_{vp} }{ s_{vp}} \\ t_{b} = \frac{30 \times {10}^{3} }{75 } = 0.4 \times {10}^{3} \\ = 400 \: k$

Final Answer: The boiling point of liquid is

$400 \: k$

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