A gas expands with temperature according to the relation `V = kT^(2//3)`. What is the work done when the temperature changes by `30^(@)C`?

Options

A. `166.2 J`
B. `136.2 J`
C. `126.2 J`
D. none of these

Answer

A

Explanation

a. `PV = RT` for 1 mol
`W = int PdV = int (RT)/(V) dV`
`V = CT^(2//3)`
`dV = (2)/(3) CT^(1//3) dT` or `(dV)/(V) = (2)/(3) (dT)/(T)`
`W = int_(T_(1))^(T_(2)) RT ((2)/(3)) (dT)/(T) = (2)/(3) R (T_(2) - T_(1)) = 166.2 J`