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Question
Give an expression for work done in an isothermal process.
Solution
The work done by the gas, W = `int_("V"_"i")^("V"_"f") "P"."dV"`
Writing pressure in terms of volume and temperature,
P = `(μ"RT")/"V"`
Substituting this value we get
W = `int_("V"_"i")^("V"_"f") (μ"RT")/"V" "dV" = μ"RT" int_("V"_"i")^("V"_"f") "dV"/"V"`
By performing the integration in equation, we get
W = `μ"RT" ln ("V"_"f"/"V"_"i")`
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