A) 14 cal
B) 6cal
C) 16 cal
D) 66 cal
Correct Answer: B
Solution :
[b] \[\Delta {{U}_{iaf}}={{Q}_{iaf}}-{{W}_{iwf}}\] \[=50-20=30cal\] For path ibf \[{{Q}_{ibf}}=\Delta {{U}_{ibf}}-{{W}_{ibf}}\] \[\Delta \,{{U}_{iaf}}=\Delta {{U}_{ibf}}\] (\[\Delta U\]path independent) \[{{Q}_{ibf}}=\Delta {{U}_{iaf}}-{{W}_{ibf}}\] \[{{W}_{ibf}}={{Q}_{ibf}}-\Delta {{U}_{iaf}}\] \[=36-60=6cal.\]You need to login to perform this action.
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