Answer to Question #100719 in Mechanics | Relativity for DU LOp

Question

A sample of gas is taken through cycle abca show
n
in the p
-
V diagram of Figure 2
. The net work done is
+2.0 J. Along path
ab, the magnitude of the work done
is 4.0 J, the energy transferred to the gas as heat is +
5
.0
J. Along path ca, the energy transferred to the gas as
heat is +3.0 J.
(a) What is the change in internal energy along path ab?
(b) How much energy is transf
erred as heat along path
bc?

Answer 1

(a) The change in internal energy along path ab can be found according to the second law of thermodynamics:

$$dQ_{ab}=dU_{ab}+W_{ab},\\dU_{ab}=dQ_{ab}-W_{ab}=5-4=1\text{ J}.$$

(b) The energy transferred as heat along path bc:

$$dQ_{ab}+dQ_{bc}+dQ_{ac}=2\text{ J},\\ 5+dQ_{bc}+3=2,\\ dQ_{bc}=-6\text{ J}.$$