Heat and Thermodynamics 5 Question 53

54. An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q1=5960J,Q2=5585J,Q3=2980J and Q4=3645J respectively. The corresponding quantities of work involved are W1=2200J,W2=825J,W3=1100J and W4 respectively.

(1994, 6M)

(a) Find the value of W4.

(b) What is the efficiency of the cycle?

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Answer:

Correct Answer: 54. (a) 765J (b) 10.82

Solution:

  1. (a) In a cyclic process ΔU=0

Therefore, Qnet =Wnet  or Q1+Q2+Q3+Q4=W1+W2+W3+W4

Hence, W4=(Q1+Q2+Q3+Q4)(W1+W2+W3)

$$ ={(5960-5585-2980+3645) $$

Extra close brace or missing open brace

or W4=765J

(b) Efficiency,

η= Total work done in the cycle  Heat absorbed (positive heat) ×100=(W1+W2+W3+W4Q1+Q4)×100=(22008251100+765)5960+3645×100=10409605×100η=10.82

NOTE

  • From energy conservation

Wnet =Qtre Q-ve  (in a cycle)

η=Wnet Qtve ×100=(Q+veQve)Q+ve×100=(1QveQ+ve)×100

In the above question

Qve=|Q2|+|Q3|=(5585+2980)J=8565J and Qtve =Q1+Q4=(5960+3645)J=9605Jη=(185659605)×100η=10.82



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