Question

When a system is taken from state $i$ to state $f$ along the path $iabf$, it is found that $Q=50$ cal and $W=20$ cal. Along the path $ibf$, $Q=36$ cal. $W$ along the path $ibf$ is:

• A. 14 cal
• B. 6 cal
• C. 16 cal
• D. 66 cal

Answer 1

The correct option is B 6 cal

From first law of Thermodynamics:
$dQ=dU+DW$
where, $dQ$ is the heat supplied to the system, $dU$ is the change in internal energy of the system and $dW$ is the work done by the system.

Path iabf:
Given:
$dQ=50kJ$
$dW=20kJ$
$\therefore dU=50-20=30kJ$

Path ibf:
$dQ=36$
Since both the processes have the same initial state(i) and final state(f), $dU$ will be same for both the processes( internal energy is a state function ).
Hence substituting the value for $dQ$ and $dU$ in the equation, $dQ=dU+dW$
$\therefore 36=30+dW$
$\therefore dW=6kJ$