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mestny [16]
3 years ago
5

A sample consisting of 2 moles He is expanded isothermally at 0 degrees from 5.0dm3 to 20.0dm3. Calculate w, q and deltaU for ea

ch of the following situations: (i) A reversible expansion of the sample. (ii) An irreversible expansion of the sample against a constant external pressure equal to the final pressure of the gas. (iii) A free expansion (against zero external pressure i.e. in a vacuum) of the sample.
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
7 0

Answer:

i) \Delta U=0 w=-6293 J q=6293 J

ii) \Delta U=0 w=-3404,52 J q=3404,52 J

ii) \Delta U=0 w=0 J q=0 J

Explanation:

As the initial and final states of the sample are the same, the ΔU of the sample is, for the three cases

\Delta U=n.C_{V}.\Delta T=0 since \Delta T=0

i)Reversibly P_{ext} =P_{sys} so w can be calculated by  

w=-n.R.T.ln(\frac{V_{f}}{V_{i}})=-2 \times 8.314\frac{J}{mol K} \times 273,15K \times ln(\frac{}{5dm^{3}})=-6293 J

and because of the first law of thermodynamics

q=-w=6293 J

ii)Irreversibly with P_{ext} =P_{f}

we can calculate P_{f} by the law of ideal gases

P_{f} =\frac{n\times R\times T}{V_{f}} =\frac{2\times 0.082\frac{dm^{3}atm}{mol K}\times 273,15K}{20dm^{3}} =2,24 atm

then w can be calculated by

w=-P_{ext} \times \Delta V=-2,24 atm \times (20-5) dm^{3} \times frac{101.325J }{atm dm^{3}=-3404,52J

and  

q=-w=3404,52J

iii)a free expansion P_{ext} so w=0 (there's no work at vaccum) and q=-w=0

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The equation for the protonation of the base pyridine is the following:

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To calculate the pH of the solution we need to use the following equation:

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C₅H₅N  +  H₂O  ⇄  C₅H₅NH⁺  +  OH⁻

0.64 - x                          x              x

After entering the values of [C₅H₅N] = 0.64-x, [C₅H₅NH⁺] = x, and [OH⁻] = x, into equation (2) we can find the concentration of OH⁻:

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