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Shweta Thakur posted an Question
May 21, 2020 • 23:10 pm 30 points
  • IIT JAM
  • Chemistry (CY)

Plz explain this expression i think it is for real gas ...

hase Equitibr he Vander walls c an gas nRT Py-nb V 4) both side of equatio by v, we get an V nRTV PVV-nb .5) wnression for P,V, and P,V, as obtained from Eq.5 into Eq.3, we obtain stituting the expre SUOSt V, V, -nb V V,-nb VV AH AE+nRT (6) V, V, -nb)-V,(V,-nb) V V NOW V,-nb V,-nb (V-nb)(V, - nb) nb(V,-V,-- nb nb (7) (V-nb(V, -nb) V, -nb V,-nb quantity Substituting Eq. 7 into Eq. 6, we have 2 1 an VV nb nb AH-AE+nkV.-nb V,-nb ..(8) Substituting for AE from internal energy change, we have 1-11 2an V 1 AH n'bR nb V,-nb Comparison of Work of Expansion of an ldeal Gas and a vander Waals Gas. We know that ifor an ideal gas, numerically .(9) WnRT In (VV) ideal and for a vander Waals gas, from work of expansion numerically T in nban ..(10) Wvdw = nRT In-nbaV, V f V>> nb, then Eq.10 reduces to RT In an 1 11) V,) Hence, numerically (12) n 2 1 1 an' (VV,) Wideal W vdw -an V V, V,V the reversible isothermal expansion of an ideal gas is greater than that for a vander Waals gas. Since Tor the expansion of a gas. VV,, it is evident from Eq. 12 that numerically, the work t sre by passi

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    Dinesh khalmaniya 1 best-answer

    this is already given in the simple language if somewhere you have any difficulty you can ask me🙏

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    Dinesh khalmaniya 1

    vander wals gases are considering as real gas so this expression is for real gas.

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