Calculate the work done during combustion of 0.138 kg of ethanol, C2H5OH(. l .) at 300 K. (Given: R=8.314JK- 1mol- 1 , molar mass of ethanol =46gmol- 1 .)

Question

Calculate the work done during combustion of 0.138 kg of ethanol, C2H5OH(. l .) at 300 K. (Given: R=8.314JK- 1mol- 1 , molar mass of ethanol =46gmol- 1 .) (A) - 7482 J (B) 7482 J (C) - 2494 J (D) 2494 J

Tardigrade Admin · Accepted Answer

C2H5OH(. l .)+3O2 (. g .) arrow 2CO2 (. g .)+3H2O(. l .) â¦(i) 0.138 kg of ethanol ≡ (138 g/46 g m o l- 1)=3 moles By multiplying equation (i) by 3, we get 3C2H5OH(. l .)+9O2 (. g .) arrow 6CO2 (. g .)+9H2O(. l .) Δ ng=6-9=3 w=-PΔ V w=-Δ ngRT=-(.-3.)× 8.314× 300 (.∵ PV=nRT.) =7482.6 J

Q. Calculate the work done during combustion of 0.138 kg of ethanol, C2​H5​OH(l)​ at 300 K. (Given: R=8.314JK−1mol−1 , molar mass of ethanol =46gmol−1 .)

- 7482 J

7482 J

- 2494 J

2494 J

C2​H5​OH(l)​+3O2(g)​→2CO2(g)​+3H2​O(l)​ …(i) 0.138 kg of ethanol ≡46gmol−1138g​=3 moles By multiplying equation (i) by 3, we get 3C2​H5​OH(l)​+9O2(g)​→6CO2(g)​+9H2​O(l)​ Δng​=6−9=3 w=−PΔV w=−Δng​RT=−(−3)×8.314×300 (∵PV=nRT) =7482.6 J

Q. 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 $= 46 g m o l^{- 1}$ .)