A sample of 2 kg of monatomic helium (assumed ideal) is taken through the process ABC another sample of 2 kg of the same gas is taken through the process ADC as shown in Fig. Given molecular mass of helium = 4.
a. What is the temperature of helium in each of the states A, B, C and D ?
b. Is there any way of telling afterwards which sample of helium went through the process ABC and which went through the process ADC ? Write yes or no.
How much is the heat involved in each of the process ABC and ADC ?

Answer

(i) `T_(A) = 120K, T_(B) = 241 K`,
`T_(C) = 481 K, T_(D) = 241 K`
(ii) No
(iii) `DeltaQ_(ABC) = (13)/(4) xx 10^(6)J; DeltaQ_(ADC) = (11)/(4) xx 10^(6)J`

Explanation

Number of gram moles of `He`,
`n = (m)/(M) = (2 xx 10^(3))/(4) = 500`
(i) `V_(A) = 10 m^(3), (i) P_(A) = 5 xx 10^(4) N//m^(2)`
`:. T_(A) = (P_(A)V_(A))/(nR) = ((10)(5xx10^(4)))/((500)(8.31))`
or `T_(A) = 120.34 K ~~ 120 K`
Similarly, `V_(B) = 10 m^(3), P_(B) = 10 xx 10^(4) N//m^(2)`
`:. T_(B) = ((10)(10xx10^(4)))/((500)(8.31))K`
`:. T_(B) = 240.68 K ~~ 241 K`
`V_(C) = 20 m^(3), P_(C) = 10 xx 10^(4) N//m^(2)`
`:. T_(C) = ((20)(10xx10^(4)))/((500)(8.31))K`
`T_(C) = 481.36 K ~~ 481 K` and `V_(D) = 20 m^(3),P_(D) = 5 xx 10^(4) N//m^(2)`
`V_(D) = ((20)(5xx10^(4)))/((500)(8.31))K`
`T_(D) = 240.68 K ~~ 241 K`.
(ii) No, if is not possible to tell afterwards which sample went through the process `ABC` or `ADC`. But we can say the process which require more work goes through process `ABC`.
(iii) In the process `ABC`,
`DeltaU = nC_(v)DeltaT = n ((3)/(2)R) (T_(C)-T_(A))`
`= (500) ((3)/(2)) 8.31 (481.36 - 120.34) J`
`DeltaU = 2.25 xx 10^(6) J` and
`DeltaW = "Area under" BC = (20 - 10) (10) xx 10^(4)J = 10^(6)J`
`:. DeltaQ_(ABC) = DeltaU +DeltaW = (2.25 xx 10^(6) +10^(6))J`
`DeltaQ_(ABC) = 3.25 xx 10^(6)J`
In the process `ADC, DeltaU` will be same (because it depends on initial and final temperature only)
`DeltaW = "Area under" AD`
`= (20 - 10) (5 xx 10^(4))J`
`= 0.5 xx 10^(6)J`
`DeltaQ_(ADC) = DeltaU + DeltaW = (2.25 xx 10^(6) +0.5 xx 10^(6))J`
`DeltaQ_(ADC) = 2.75 xx 10^(6)J`.