NEET Physics Thermodynamics Questions Solved

NEET - 2015

Figure below shows two paths that may be taken by a gas to go from a state A to a state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be-


(a) 380 J 
(b) 500 J
(c) 460 J
(d) 300 J

Since, initial and final points are same
                         

So, UABC = UAC                                                ........iAlso AB is isochoric processSo   WAB=0and   Q=U+WSo,   QAB=UAB=400 JNext  BC is isobaric process.So,   QBC=UBC+WBC                       = UBC+PVBC     100=UBC+6×104 2×10-3      UBC=100-120 =-20 JFrom Equation (i):       UABC=UAC UAB+UBC = QAC-WAC 400+-20=QAC-Area under AC QAC=380+2×104×2×10-3+12×2×10-3×4×104                   =380+40+40    QAC=460 J

Difficulty Level:

  • 13%
  • 34%
  • 45%
  • 10%
Crack NEET with Online Course - Free Trial (Offer Valid Till August 25, 2019)