$ 1 \mathrm{Mole} $ of $ \mathrm{CO}_{2} $ gas at $ 300 \mathrm{~K} $ expanded under the reversible adiabatic condition such that its volume becomes $ 27 $ times. The magnitude of work done (in $ \mathrm{kJ} / \mathrm{mol} $ ) is: (Given $ \gamma=1.33 $ and $ \mathrm{C}_{\mathrm{v}}=25.10 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1} $ for $ \mathrm{CO}_{2} $ ) report your answer by rounding it up to nearest whole number

Question

$ 1 \mathrm{Mole} $ of $ \mathrm{CO}_{2} $ gas at $ 300 \mathrm{~K} $ expanded under the reversible adiabatic condition such that its volume becomes $ 27 $ times. The magnitude of work done (in $ \mathrm{kJ} / \mathrm{mol} $ ) is: (Given $ \gamma=1.33 $ and $ \mathrm{C}_{\mathrm{v}}=25.10 \mathrm{Jmol}^{-1} \mathrm{~K}^{-1} $ for $ \mathrm{CO}_{2} $ ) report your answer by rounding it up to nearest whole number,

MARKS App · Accepted Answer

Number of moles = 1 T 1 = 300 K V 2 = 27 V 1 T V 1 γ - 1 = T 2 V 2 γ - 1 T 1 T 2 = V 2 V 1 γ – 1 T 2 = 300 1 27 1 3 T 2 = 100 K Adiabatic condition, q = 0 ; Δ E = w = n C v T 2 – T 1 w = 1 × 25.10 100 – 300 = - 5020 J / m o l e w = - 5.02 k J / m o l e

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