Q.

An ideal gas undergoes a cyclic process as shown in Figure

Heat absorbed by the system during the process CA is

 Question Image
ΔUBC=5kJmol1 
qAB=2kJmol1 
WAB=5kJmol1 
WCA=3kJmol1

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a

5 kJ mol1

b

18 kJ mol1

c

+5 kJ mol1

d

+18 kJ mol1

answer is B.

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Detailed Solution

For AB process 

EAB=qAB+wAB=2-5=-3

For cyclic processes, change in internal energy is zero.

EAB+EBC+ECA=0

ECA=-(-3)-(-5)=8

ECA-wCA=qCA

qCA=8-3=5

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