Question

When a system is taken from thermodynamic state i to $f$ along the path iaf (see figure), it is found that the heat $Q$ absorbed by the system is 50 cal and work $W$ done by the system is equal to 20 cal. Along the path ibf $Q=36$ cal. What is W along the path ibf?

• A. 6 cal
• B. 18 cal
• C. $40cal$
• D. -43 cal

Answer 1

Answer: (A)

From the first law of thermodynamics
$\Delta U=U_f-U_i=Q-W\dots$(i)
Where $Q$ is the heat absorbed by the system and $W$ is the work done by the system.
For the path iaf; we have
$Q=+50cal,W=+20cal$
$\therefore \Delta U=Q-W=50-20=30cal$
Since internal energy of the system is a function of its thermodynamic state and depends only on temperature and no other factor. Therefore $\Delta U$ between i to $f$ is always 30 cal whatever path be adopted.
For the path ibf,
$Q=+36cal\text{ and }\Delta U=30cal$
$\therefore W=Q-\Delta U=36-30=6cal$