Two moles of an ideal monoatomic gas undergo a cyclic process which is indicated on a `P -U` diagram, where `U` is the internal energy of the gas. The work done by the gas in the cycle is `k xx 10^(2) In2`. Find `k`.

Answer

12

Explanation

Change in internal energy in process
`Delta U_(AB) = 3600 - 1800 = 1800 cal = (f)/(2) n R DeltaT_(AB)`
`DeltaU_(BC) = 0`
`DeltaU_(CD) =- 1800 cal = (f)/(2) nRT DeltaT_(CD)`
`Delta U_(DA) = 0`
work done by gas at constant pressure is `nRT DeltaT`
Therefore `W_(AB) = nRT Delta T_(AB) = (2DeltaU_(AB))/(f) = 3600//f`
`[DeltaU = (nfRDeltaT)/(2)] = 1200 cal`
and `W_(CD) = nR DeltaT_(CD) = (2DeltaU_(CD))/(f) = - 3600//f`
`=- 1200 cal`
work done by gas at constant temperature
`W = nRT In ((p_(i))/(p_(f))) = (2U)/(f) In ((P_(i))/(P_(F))) [U = (nfRT)/(2)]`
`W_(DA) = (2xx 1800)/(3) ln ((2)/(4)) =- 1200 ln 2`
`W_(BC) = (2 xx 3600)/(3) ln ((4)/(2)) = 2400 ln 2`
Total work done by gas `= W_(AB) +W_(BC) +W_(CD) +W_(DA) = 1200 ln2`