Question

 Two moles of helium gas are taken over the ABCDA, as shown in the P-T diagram 

(i) Assuming the gas to be ideal the work done by the gas in taking it from A to B is 

(A) 200 R (B) 300 R (C) 400 R (D) 500 R

(2) The work done on the gas in taking it from D to A is- 

(A) -414R (B) +414 R (C) -690 R (D) +690 R

(3) The work done on the gas in the cycle ABCDA is- 

(A) Zero (B) 276 R (C) 1076 R (D) 1904 R

Answer 1

Correct option (i)(C (ii) (B) (iii) (B)

Explanation:

(i) w A–B –w = PC∆v = nR∆T

 2 × 10⁵ × T

= 2 × 8.314 × 200 = 400 R

(ii) Work done= – nRT ln V₂/V₁

(iii) Net work done in cycle

∴ Total mole = 2 R (500 – 300) –2R 500 λ ln 2

= 2R (500 – 300) + 2 × 300 λ ln 2 

= 2R (300 – 500) ln

= 420 R ln 2

= 276 R