2 moles of He gas (γ = 5/3) of 20 lit. volume at 27 0 C in subjected to constant pressure is expanded to double its volume. The work done in isobaric process is –
Text Solution
Verified by ExpertsThe correct answer is:
A
For isobaric expansion
= 
T 1 = 27 + 273 = 300 K.
V 2 = 2V 1
T 2 = 
× T 1
µ = 2 mole
T 2 = 
× 300 = 600 K
W isobaric = P (V 2 – V 1 ) = µR (T 2 – T 1 )
= 2 × 8.3 (600 – 300)
= 2 × 8.3 × 300 = 4980 J
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