Calculate the work done during combustion of 0.138 kg of ethanol, C 2 H 5 O H ( l ) at 300 K. (Given: R = 8.314 J K - 1 m o l - 1 , molar mass of ethanol = 46 g m o l - 1 .)

Question

Calculate the work done during combustion of 0.138 kg of ethanol, C 2 H 5 O H ( l ) at 300 K. (Given: R = 8.314 J K - 1 m o l - 1 , molar mass of ethanol = 46 g m o l - 1 .), – 7482 J, 7482 J, – 2494 J, 2494 J

MARKS App · Accepted Answer

C 2 H 5 O H ( l ) + 3 O 2 ( g ) → 2 C O 2 ( g ) + 3 H 2 O ( l ) …(i) 0.138 kg of ethanol ≡ 138 g 46 g m o l - 1 = 3 moles By multiplying equation (i) by 3, we get 3 C 2 H 5 O H ( l ) + 9 O 2 ( g ) → 6 C O 2 ( g ) + 9 H 2 O ( l ) Δ n g = 6 - 9 = 3 w = - P Δ V w = - Δ n g R T = - ( - 3 ) × 8.314 × 300 ( ∵ P V = n R T ) = 7482.6 J

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