Question

When a system is taken from state A to state B along the path ACB, 80J of heat flows into the system and the system does 30J of work.
(A) How much heat flows into the system along path ADB if the work done is 10J?
(B) When the system is returned from state B to A along the curved path, the work done on the system is 20J. 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,

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

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