First slide
First law of thermodynamics
Question

A thermodynamical process is shown in the figure (4). The pressures and volumes corresponding to some points in the fig. (4) are

PA=3×104PaVA=2×103m3PD=8×104PaVD=5×103m3
In the process AB,600 J of heat is added to the system and in process 8C,200 | of heat is added to the system. The change in internal energy of the system in process
AC would be

 

Moderate
Solution

For process AB, 
ΔQ=ΔU+ΔW=UBUA+dW 600=UBUA+0 or UBUA=600
For process BC,
ΔQ=ΔU+dW200=UCUB+PBVCVB=UCUB+8×1045×1032×103=UCUB+240 UCUB=200240=40J
For process AC,
Change in internal energy=UCUA
=UCUB+UBUA=40+600=560J

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