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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 ? |
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Answer» Correct 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` 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`. |
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