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 each 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.

Respuesta :

Answer:

i) [tex]\Delta U=0 w=-6293 J q=6293 J[/tex]

ii) [tex]\Delta U=0 w=-3404,52 J q=3404,52 J[/tex]

ii) [tex]\Delta U=0 w=0 J q=0 J[/tex]

Explanation:

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

[tex]\Delta U=n.C_{V}.\Delta T=0[/tex] since [tex]\Delta T=0[/tex]

i)Reversibly [tex]P_{ext} =P_{sys}[/tex] so [tex]w[/tex] can be calculated by  

[tex]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[/tex]

and because of the first law of thermodynamics

[tex]q=-w=6293 J[/tex]

ii)Irreversibly with [tex]P_{ext} =P_{f}[/tex]

we can calculate [tex]P_{f}[/tex] by the law of ideal gases

[tex]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[/tex]

then w can be calculated by

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

and  

[tex]q=-w=3404,52J[/tex]

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

ACCESS MORE
EDU ACCESS
Universidad de Mexico