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]