Question

2 moles of He gas (g = 5/3) of 20 lit. volume at 27⁰C in subjected to constant pressure is expanded to double its volume. The work done in isobaric process is –

• A. 4980 J
• B. 7470 J
• C. 2490 J
• D. 11454 J

Answer 1

Answer: (A)

4980 J

Answer 2

For isobaric expansion

$\frac{V_{1}}{T_{1}}$=$\frac{V_{2}}{T_{2}}$ T₁ = 27 + 273 = 300 K.

V₂ = 2V₁

T₂ =$\frac{V_{2}}{V_{1}}$× T₁

µ = 2 mole

T₂ =$\frac{2V_{1}}{V}$× 300 = 600 K

W_isobaric = P(V₂ – V₁) = µR(T₂ – T₁)

= 2 × 8.3 (600 – 300)

= 2 × 8.3 × 300 = 4980 J

W_{isobaric} = 4980J \end{matrix}$$