Question #100723
A sample of gas is taken through cycle abca shown 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 transferred as heat along path bc?
1
Expert's answer
2019-12-24T14:29:41-0500


a)


Uab=QabWabU_{ab}=Q_{ab}-W_{ab}

Uab=54=1.0 JU_{ab}=5-4=1.0\ J

b)


Qbc=QabcaQabQcaQ_{bc}=Q_{abca}-Q_{ab}-Q_{ca}

Qbc=Wabca(WabUab)QcaQ_{bc}=W_{abca}-(W_{ab}-U_{ab})-Q_{ca}

Qbc=2(41)3=4 JQ_{bc}=2-(4-1)-3=-4\ J


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