Q. 34.0( 2 Votes )

Figure shows two process A and B on a system. Let ΔQ1 and ΔQ2 be the heat given to the system in processes A and B respectively. Then

A. ΔQ1 > ΔQ2

B. ΔQ1 = ΔQ2

C. ΔQ1 < ΔQ2

D. ΔQ1< ΔQ2

Answer :

Initial and final points of both processes A and B are same. Therefore, internal energy in both the processes will be the same because internal energy is a state variable, independent of the path taken.

The area under the P-V curve gives the work done on the system. From the graph, it can be seen the area under the curve for process A is more than the area under the curve for process B., therefore, work done on the system in process A ΔW1 is more than work done on the system in process B ΔW2

ΔW1 > ΔW2 …..(i)

According to First law of thermodynamics,


Where ΔQ=heat supplied/extracted to/from the system

ΔU=change in internal energy

ΔW=work done by/on the system

For process A

ΔQ1=ΔU+ΔW1 …(ii)

For process B

ΔQ2=ΔU+ΔW2 …(iii)

From equation (i) ,(ii) and (iii) it is clear that ΔQ1 > ΔQ2.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
JEE Advanced Level Questions of NLM48 mins
Work done in Thermodynamic ProcessFREE Class
Microsporogenesis & Megasporogenesis49 mins
NEET 2021 | Transcription - An important Topic62 mins
DNA Fingerprinting42 mins
Meselson and Stahl experiment49 mins
Interactive Quiz on Molecular basis of Inheritance-02FREE Class
MCQs of Ecology for NEET52 mins
Interactive Quiz on Sexual Reproduction in flowering plants44 mins
Pedigree chart58 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses