Discussion Forum



1)

Calculate the work done during combustion of 0.138 kg of ethanol, C2H5OH (l) at 300K.

Given: R=8.314 J K-1 mol-1, the molar mass of ethanol=46  g mol-1.


A) -7482 J

B) 7482 J

C) -2494 J

D) 2494 J

Answer:

Option B

Explanation:

The combustion of ethanol , involves the following reaction, 

$C_{2}H_{5}OH(l)+3O_{2}(g)\rightarrow 2CO_{2}+3H_{2}O$

 Given, Mass of ethanol= 0.138 kg= 138 g

Temperature =300K

  R= 8.314 J K-1 mol-1

  Molar  mass of ethanol =46 g mol-1

 No. of moles of ethanol = Mass of ethanol / Molar mass of ethanol

    = $\frac{138}{46}$=3

 Work done (W)  during combustion of 0.138 kg of  C2H5OH= nRT

       W=nRT

 W= 3 x 8.314 JK-1 mol-1 x 300 K

     W= 7482.6J

        =7482 J