Question

When a thermodynamic system is taken from an initial state I to a final state F along the path IAF, as shown in Fig. the heat energy absorbed by the system is `Q = 55 J` and work done by the system is `W = 25 J`. If the same system is taken along the path IBF, the value of `Q = 35 J`.
(a) If `W = - 15 J` for the curved path FI, how much heat energy is lost by the system along this path ?
(b) If `U_(1) = 10 J`, what is `U_(F)` ?
(d) If `U_(B) = 20 J`, what is Q for the process BF and IB ?

Answer 1

The first law of thermodynamics states that
`Delta Q = Delta U + Delta W` or `Q = (U_(F) - U_(I)) + W`
Here `U_(I)` and `U_(F)` are the internal energies in the initial and the final state, respectively. Given that for path `IAF, Q = 55 J` and `W = 25 J`. Therefore,
`Delta U = U_(F) - U_(I) = Q - W = 55 - 25 = 30 J`
The internal energy is independent of the path, it depends only on the initial and final states of the system. Thus, the internal energy between I and F states is 30 J irrespective of the path followed by the system.
a. For path IBF, `Q = 35 J` and `Delta U = 30 J`. Therefore, `W = Q - Delta U = 35 - 30 = 5 J`
b. `W = - 15 J`, but `Delta U = - 30 J`. Therefore, `Q = W + Delta U = - 15 - 30 = - 45 J`.
Given `U_(I) = 10 J`. Therefore, `U_(F) = Delta U + U_(I) = 30 + 10 = 40 J`.
d. The process BF is isochoric, i.e., the volume is constant. Hence,
W = 0. Therefore, `Q = (Delta U)_(BF) = U_(F) - U_(B) = 40 - 20 = 20 J`.
The process IB is isobaric (constant pressure). Therefore.
`Q = (Q)_(IBF) - (Q)_(BF) = 35 - 20 = 15 J`.