Q.
A system is taken fromstate $a$ to state $c$ by two paths $a d c$ and $a b c$ as shown in the figure. The internal energy at $a$ is $U_{a}=10 \,J$. Along the path $a d c$ the amount of heat heat absorbed $d Q_{1}=50 \,J$ and the work obtained $d W_{1}=20 J$ whereas along the path $a b c$ the heat absorbed $d Q_{2}=36 J$. The amount of work along the path $a b c$ is
Solution:
According to first law of thermodynamics,
$\delta Q=\delta U+\delta W$
Along the path adc Change in internal energy,
$\delta U_{1}=\delta Q_{1}-\delta W_{1}$
$=50 J-20 J=30 J$
Along the path $abc$
Change in internal energy,
$\delta U_{2}=\delta Q_{2}-\delta W_{2} $
$\delta U_{2}=36 J-\delta W_{2}$
As change in internal energy is path independent.
$\therefore \delta U_{1}=\delta U_{2} $
$\therefore 30 J=36 J-\delta W_{2}$
$\delta W_{2}=36 J-30 J=6 J$
