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

Question

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

Tardigrade Admin · Accepted Answer

The combustion of ethanol, involves the following reaction.

C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} + 3 H_{2} O

Given,

Mass of ethanol

= 0.138 k g = 138 g

Temperature =

300 K

R = 8.314 J K^{- 1} m o l^{- 1}

Molar mass of ethanol

= 46 g m o l^{- 1}

No. of moles of ethanol =

\frac{Mass of ethanol}{Molar mass of ethanol}

= \frac{138}{46} = 3

Work done

(W)

during combustion of

0138 k g

of

C_{2} H_{5} O H = n RT

W = n RT

W = 3 \times 8.314 J K^{- 1} m o r^{- 1} \times 300 K

W = 7482.6 J

\approx 7482 J

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

-7482 J

7482 J

-2494 J

2494 J

Q. Calculate the work done during combustion of $0.138 k g$ 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 = $46 g m o l^{- 1}$ .