Question

A lab sample of gas is taken through cycle abca .The net work done on the gas is -1.2J.Along path ab,the change in internal energy is +3.0J and the work done on the gas is -5.0J.Along path ca,the energy transferred to the gas as heat is +2.5J.

(a). Determine the energy transferred to the gas as heat along path ab.

(b). Determine the energy transferred to the gas as heat along path bc.

Answer 1

$∆W_{ca} = -1.2\textsf{ J}\\ ∆U_{ab} = +3.0\textsf{ J}\\ ∆W_{ab} = -5.0\textsf{ J}\\ Q_{ca} = +2.5\textsf{ J}\\$

According to the first law of thermodynamics;

$ΔU_{ca} = Q_{ca} − ∆W_{ca}\\ +2.5= Q_{ca}-(-1.2)\\ 2.5 = Q_{ca}+1.2\\ Q_{ca} = 2.5-1.2\\ Q_{ca} = 1.3\textsf{ J}$

(a)

According to the first law of thermodynamics;

$Q_{ab} =\ ?\\ \quad\\ ΔU_{ab} = Q_{ab} − ∆W_{ab}\\ 3.0 = Q_{ab} -(-5.0)\\ 3.0 = Q_{ab} + 5.0\\ Q_{ab}= 3.0- 5.0 = -2.0\textsf{ J}\\$

(b)

Heat is a path function

$Q_{bc} =\ ?\\ \quad\\ 1.3\textsf{ J}-2.0\textsf{ J} + Q_{bc} = 0\\ Q_{bc} = 0.7\textsf{ J}$