Question

When a system is taken from state A to state B along the path ACB, 80 J of heat glows into the system and the system does 30 J of work, (a) How much heat flows into the system along path ADB if the work done is 10 J ? (b) When the system is returned from state B to A along the curved path, the work done on the system is 20 J. Does the system absorb or liberate heat and how much ? (c) If $U_D-U_A=40j$, find the heat absorbed in the process AD and DB.

Answer 1

Since $\Delta U$ is independent of path.

$\therefore \Delta U(A\to B)=q\left(ACB\right)+w\left(ACB\right)=80-30=50J$

(a) $\Delta U=50J=q\left(ADB\right)+w\left(ADB\right)$

$\Rightarrow q\left(ADB\right)=50-\left(10\right)=40J$

(b) $q(B\to A)=\Delta U(B\to A)-w(B\to A)=(-50)-20=-70J[\because \Delta U(A\to B)=-\Delta U(B\to A)]$

Hence the system liberates heat.

(c) $\Delta U\left(ADB\right)=\Delta U(A\to D)+\Delta U(D\to B)$

$\Rightarrow \Delta U(D\to B)=50-40=10J$

Again,

$\Delta U(D\to B)=q(D\to B)+w(D\to B)$

$\because V(D\to B)$ is constant.

$\therefore w(D\to B)=0$

$\therefore q(D\to B)=\Delta U(D\to B)=10J$

Now,

$q\left(ADB\right)=q(A\to D)+q(D\to B)$

$\Rightarrow q(A\to D)=40-10=30J[\because q\left(ADB\right)=40J]$