A litre of dry air at STP allowed to expand to a volume of 3 L under adiabatic conditions. If γ = 1.40 , the work done is ( 3 1.4 = 4.6555 )

Step-by-step solution

Here, V 1 = 1 L = 10 - 3 m 3 , V 2 = 3 L = 3 × 10 - 3 m 3 p 1 = 1 atm = 1.013 × 10 5 N m - 2 , γ = 1.40 , W = ? As changes are adiabatic, p 1 V 1 γ = p 2 V 2 γ p 1 p 2 = V 2 V 1 γ = 3 1.4 = 4.6555 ∴ p 2 = p 1 4.6555 = 1.013 × 10 5 4.6555 = 0.217 × 10 5 N m - 2 Work done, W = p 1 V 1 - p 2 V 2 γ - 1 W= 1.013 × 10 5 × 10 - 3 - 0.217 × 10 5 × 3 × 10 - 3 1.4 - 1 W = 90.5   J