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 ?

Explanation

Given that mass of helium used in the process is m = 2 kg, number of moles can be given as
`n = (m)/(M) = (2)/((2 xx 10^(-3))) = 500`
At different states, from Fig. the pressure and volume fo gas are also given as
`P_(A) = P_(D) = 10^(4) N//m^(2)`
`P_(B) = P_(C ) = 10^(5) N//m^(2)`
`V_(A) = V_(B) = 10 m^(3)`
`V_(C ) = V_(D) = 20 m^(2)`
a. From gas law, we have
`T_(A) = ((P_(A) V_(A)))/((nR)) = ((5 xx 10^(4) xx 10))/((500 xx 8.314)) = 120.3 K`
`T_(B) = ((P_(B) V_(B)))/((nR)) = ((10^(5) xx 20))/((500 xx 8.314)) = 481.11 K`
`T_(D) = ((P_(D) V_(D)))/((nR)) = ((5 xx 10^(4) xx 20))/((500 xx 8.314)) = 120.3 K`
b. Since the gas is taken from same initial state to same final state C, no matters whatever be the path, the answer is no
c. In process ABC, the change in internal energy is
`Delta U_(ABC) = U_(C ) - U_(A) = (f//2) nRT (T_(C)) - T_(A))`
`= (3//2) xx 500 xx 8.314 (481.11 - 120.3) = 2.25 xx 10^(6) J`
Net done in process ABC is
`W_(ABC) = W_(AB) + W_(BC)`
= 0 area below curve BC
`= 0 + 10^(5) xx 10`
`= 10^(6) J`
Thus, from the first law of thermodynamic, heat supplied in process ABC is
`Q = W + Delta U`
or `Q = 10^(6) + 2.25 xx 10^(6)`
`= 3.25 xx 10^(6) J`
Similarly, in process ADC as being a state function, change in internal energy remains same as initial and final states are same. Thus
`Delta U_(ADC) = 2.25 xx 10^(6) J`
Thus, work done by the gas in process ADC is
`W_(ADC) = W_(AD) + W_(DC)`
= area below curve AD + 0
`= 5 xx 10^(4) xx 10`
`= 0.5 xx 10^(6) J`
Thus, from the first law of thermodynamics, heat supplied the process ADC is given as
`Q = W + Delta U`
or `= 0.5 xx 10^(6) + 2.25 xx 10^(6)`
`= 2.75 xx 10^(6) J`