Question

An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are Q_1=5960J, Q_2=-5585J, Q_3=-2980J and Q_4=3645J respectively. The corresponding quantities of work involved are W_1=2200J, W_2=-825J, W_3=-1100J and W_4 respectively. (a) Find the value of W_4. (b) What are the efficiency of the cycle?

Answer 1

Hence the value of W4 is 765 J and the efficiency of the cycle is η = 10.82 %

Explanation:

(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)-(2200-825-1100)}

or W4 = 765 J

(b) Efficiency:

η = Total work done in the cycle / Heat absorbed (positive heat)×100

=(W1+W2+W3+W4 / Q1 + Q4) × 100

= {(2200−825−1100+765) / 5960+3645}×100

= 1040 / 9605×100

η = 10.82 %

Hence the value of W4 is 765 J and the efficiency of the cycle is η = 10.82 %

Answer 2

Explanation:

n=10.82%

W4=765 J

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