Give the relationsp between isothermal and free expansion of an ideal gas chemstry
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Free expansion of a gasoccurs when it is subjected to expansion in a vacuum (pex=0). During free expansion of an ideal gas, the work done is 0 be it a reversible or irreversibleprocess.It is known that the change in internal energy of a system is given as:∆U = q + w—(1)Where ∆U represents the change in internal energy, q is the heat given by the system and w is the work done on the system.Depending upon the type of process the above equation can be written in different ways.The work done in vacuum, w = pex∆V. Therefore equation 1 can be given as:∆U = q + pex∆VIf this process is done at constant volume then ∆V = 0. Thus,∆U = Qqvimplies that the heat is supplied at a constant volume.When an ideal gas is subjected to isothermal expansion (∆T = 0) in vacuum the work done w = 0 as pex=0. As determined by Joule experimentally q =0, thus ∆U = 0.For isothermal reversible and irreversible changes; equation 1 can be expressed as:Isothermal reversible change: q = -w = pex(Vf-Vi)Isothermal reversible change: q = -w = nRTln (Vf/Vi) = 2.303 nRT log (Vf/Vi)Adiabatic change: q =0, ∆U = wad
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Answer:nRTln(V2/V1)
Explanation:
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