A sample of 2 kg monoatomic helium (assumed ideal) is taken through the process and another sample of 2kg of the same gas is taken through the process (see figure). Given molecular mass of helium = 4

How much is the heat involved in the process (in kJ)?

Answer

(i) `T_(A) = 120.34K, T_(B) =240.68K , T_(C) = 481.36K, T_(D) = 240.68K` (ii) No (iii) ` Q_(ABC) = 3.25 xx 10^(6)J , Q_(ADC) = 2.75 xx 10^(6)J`

Explanation

(i) `T_(A) =(P_(A)V_(A))/(nR) = (5 xx 10^(4) xx 10)/(((2000)/(4)) xx 8.314) = 120.3^(@)K`
`T_(B) = (P_(B) V_(B))/(nR) = 2T_(A) = 240.6 ^(@)K`
`T_(C) = (P_(C)V_(C))/(nR) = 2T_(B) = 481.3^(@)K`
`T_(D) = (P_(D)V_(D))/(nR)= T_(B) = 240.6^(@)K`
(ii) No. we cant not predict the direction of reach
(iii) Process `ABC` : `W = P DeltaV = 10 xx 10^(4) xx (20-10)= 10^(6)J`
`DeltaU = nC_(V)DeltaT = ((2000)/(4)) ((3R)/(2)) xx (T_(C)-T_(A)) = 2.25 xx 10^(6) J therefore Q = 3.25 xx 10^(6)J`
Process `ADC rarr`
`W = 5 xx 10^(4) xxx (20-10) = 0.5 xx 10^(6)J`
`DeltaU = nC_(V)(T_(C) -T_(A)) = 2.25 xx 10^(6)J`
`therefore Q = 2.75 xx 10^(6)J`