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

2750

Explanation

Number of gram moles of He
`n = m/M = (2xx10^3)/4 = 500`
No, it is not possible to tell afterwards which sample went through the process ABC or ADC. But during the processes, then the process which require more work goes through process
(c) In the process ABC
`Delta U = n C_V Delta T = n (3/2R) (T_C - T_A)`
`=(500)(3/2)(8.31)(481.36 -120.34)j`
`Delta U = 2.25 xx10^6 J`
and `Delta W` = Area under
`= (20-10)(10)xx10^4 J = 10^6 J" ":." "Delta Q_("ABC") = Delta U + Delta W = (2.25 xx10^6 + 10^6)J`
`Delta Q_("ABC") = 3.25 xx10^6 J`
In the process `"ADC "Delta U` will be same (because it depends on initial and final temperatures only)
`Delta W` = Area under AD
` = (20-10)(5xx10^4) J = 0.5 xx10^6 j`
`Delta Q_("ADC") = Delta U + Delta W = (2.25xx10^6 + 0.5 xx10^6)J" , "Delta Q_("ADC") = 2.75xx10^6 J`