For a sample of perfect gas when its pressure is changed isothermally from `p_(i)` to `p_(f)` , the entropy change is given byA. `DeltaS= nRln((p_(f))/( p_(i)))`B. `DeltaS= nRln((p_(i))/( p_(f)))`C. `DeltaS= nRTln((p_(f))/( p_(i)))`D. `DeltaS= nRTln((p_(i))/( p_(f)))`

For a sample of perfect gas when its pressure is changed isothermally

1.	For a sample of perfect gas when its pressure is changed isothermally from `p_(i)` to `p_(f)` , the entropy change is given byA. `DeltaS= nRln((p_(f))/( p_(i)))`B. `DeltaS= nRln((p_(i))/( p_(f)))`C. `DeltaS= nRTln((p_(f))/( p_(i)))`D. `DeltaS= nRTln((p_(i))/( p_(f)))`
Answer» Correct Answer - b For n moles of an ideal gas undergoing reversible expansion when temperature changes from`T_(1)` to`T_(2)`and pressure changes from `p_(1)` to `p_(2)`, entropy change is given by `DeltaS = n C_(p) ln. (T_(2))/(T_(1)) + nRln. (P_(1))/(P_(2)) ` or `DeltaS = nC_(p)ln. (T_(f))/(T_(i))+ nRTln. (p_(i))/(p_(f))` For isothermal expansion, `T_(i) = T_(f)` so that `ln. (T_(f))/(T_(i)) = ln 1= 0`. Hence, `DeltaS = nR ln. (p_(i))/( p_(f))`

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For a sample of perfect gas when its pressure is changed isothermally from `p_(i)` to `p_(f)` , the entropy change is given byA. `DeltaS= nRln((p_(f))/( p_(i)))`B. `DeltaS= nRln((p_(i))/( p_(f)))`C. `DeltaS= nRTln((p_(f))/( p_(i)))`D. `DeltaS= nRTln((p_(i))/( p_(f)))`

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