Consider a vertical tube open at both ends. The tube consistss of two parts, each of different cross sections and each part having a piston which can move smoothly in respective tubes. The two piston which can move smoothly in respective tube wire. The piston are joined together by an inextensible wire. The combined mass of the two piston is `5 kg`and area of cross section of the upper piston is `10 cm^(2)` greater than that of the lower piston. Amount of gas enclosed by the pistons is `1 mol`. When the gas is heated slowly, pistons move by `50 cm`. Find the rise in the temperature of the gas in the form `X//R K`, where `R` is universal gas constant. Use `g = 10 m//s^(2)` and outside pressure `= 10^(5) N//m^(2))`.

Answer

75

Explanation

Due to the heating pressure inside is constant. Let inside pressure be `P`. Then for equilibrium of the system:
`P(A_(1) -A_(2)) = P_(0) (A_(1) -A_(2)) +(m_(1)+m_(2)) g`
`rArr P DeltaA = P_(0) DeltaA +mg`
`rArr P DeltaV = (P_(0)DeltaA +mg)l`
`l` is displacement of the pistons.
`P. DeltaV = nRTDeltaT`.
`DeltaT = (PDeltaV)/(nR) = ((P_(0)DeltaA+mg)l)/(nR)`

`=((10^(5)xx10^(-3)+5xx10)(50xx10^(-2)))/(1xxR)`
`DeltaT = (75)/(R )K :. X = 75`