Question

Two moles of Helium gas undergo a reversible cyclic process as shown in figure. Assuming gas to be ideal, what is the net work involved in the cyclic process ?

• A. – 100 R$\mathcal{l}$n4
• B. +100R$\mathcal{l}$n4
• C. +200R$\mathcal{l}$n4
• D. –200R$\mathcal{l}$n4

Answer 1

Answer: (A)

– 100 R$\mathcal{l}$n4

Answer 2

Wnet = area enclosed

V = $\frac{nRT}{P}$

VA = $\frac{2R \times 300}{1} = \frac{600R}{1}$

VB = $\frac{2R \times 300}{2} = \frac{300R}{1}$; Vc = $\frac{2R \times 400}{2} = \frac{400R}{1}$

VD = $\frac{800R}{1}$

WAB = –nRT TAln $\frac{V_{B}}{V_{A}}$= –2R (300)$\frac{1}{2}$ln= 600 Rln2

WBC = –2(400 –300) R= – 200 R

WCD = – 2 R(400)ln $\frac{V_{D}}{V_{C}}$= – 800 Rln2

WAD = –1(600 R – 800 R) = 200 R

W Total dqy = WAB + WBC + WCD + WAD =– 200 Rln2 = – 100 Rln4