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QJEEM07Ph23c3: JEE Main 2007 Physics - Class 11 JEEM07Ph23c3 Solution: When a system is taken from state to state along the path iaf,…

When a system is taken from state ii to state ff along the path iaf, it is found that Q=50Q=50 cal and W=20W=20 calcal . Along the path ibfibf Q=36Q=36 cal.cal. WW along the path ibfibf is AIEEE 2007 Physics - Heat and Thermodynamics Question 378 English

Multiple Choice Options

  • A 1414 calcal
  • B 66 calcal
  • C 1616 calcal
  • D 6666 calcal

Step-by-Step Verified Solution

For path iaf, ΔU=Q−W=50−20=30 cal.\Delta U = Q - W = 50 - 20 = 30\,cal.
For path ibf, W=Q−ΔU=36−30=6 cal.W = Q - \Delta U = 36 - 30 = 6\,cal.

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