A litre of dry air at STP expands adiabatically to a volume of 3 litres. If γ = 1.40, the work done by air is: (31.4 = 4.6555) [Take air to be an ideal gas] (a) 60.7 J (b) 100.8 J (c) 90.5 J (d) 48 J

Answer: (c). P₁ = 1.013×10⁵ Pa, V₁ = 10⁻³ m ³, V₂ = 3×10⁻³ m ³. P₂ = P₁(1/3)1.4 = (P₁)/(4.6555), so P₂V₂ = P₁V₁ × (3)/(4.6555) = 0.644 P₁V₁. W = (P₁V₁ - P₂V₂)/(γ - 1) = (101.3 × (1 - 0.644))/(0.4) ≈ 90 J So the answer is (c) 90.5 J.

A litre of dry air at STP expands adiabatically to a volume of 3… | JEE Main 2020 (7 Jan, Shift 1)

Q206JEE Main 2020 · 7 JAN · SHIFT 1Medium

Thermodynamic Processes

A litre of dry air at STP expands adiabatically to a volume of 3 litres. If $\gamma$ = 1.40, the work done by air is: (31.4 = 4.6555) [Take air to be an ideal gas]

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Answer: (c)

$P_1 = 1.013\times10^5$ Pa, $V_1 = 10^{-3}$ m $^3$ , $V_2 = 3\times10^{-3}$ m $^3$ .
$P_2 = P_1\left(\frac{1}{3}\right)^{1.4} = \frac{P_1}{4.6555}$ , so $P_2V_2 = P_1V_1 \times \frac{3}{4.6555} = 0.644\,P_1V_1$ .
$W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{101.3 \times (1 - 0.644)}{0.4} \approx 90$ J
So the answer is (c) 90.5 J.

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