Q4 / 160 · NTA Abhyas (JEE Main) 2020 · Physics · Thermodynamics

An ideal gas undergoes the cyclic process abca as shown in the figure. The internal energy change of the gas along the path ca  is  −180 J . The gas absorb 250 J of heat along the path ab  and also absorbs  60 J  of heat along the path bc . Find the work done by the gas along the path abc

Figure

  • Correct answer
    A
    130 J
  • B
    140 J
  • C
    120 J
  • D
    100 J

Solution

Change in internal energy for a cyclic process is zero. Thus,

∆Uabc+∆Uca=0∆Uabc=-∆Uca=180 J
ΔQ=ΔW+ΔU

250+60=ΔW+180

ΔW=130 J

Hence, the correct answer is option: 'A'.

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