Question

The work done in the dm3 bar when 200 mL of ethylene gas and 150 mL of HCl gas were allowed to
at I bar pressure is....
​

Answer 1

15.0 J
Explanation:
W = - Pext ∆V = - Pext (V2 - V1 )
According to the equation of reaction,
1 mole of C2 H4 reacts with 1 mole of HCl to produce 1 mole of C2 H5 Cl. Hence, 150 mL
of HCl would react with only 150 mL of C2 H4 to produce 150 mL of C2 H5 Cl.
V1 = 150 mL + 150 mL = 300 mL = 0.3 dm3
V2 = 150 mL = 0.15 L, Pext = 1 bar
Substitution of these quantities in above
W = -1 bar (0.15 dm^3 - 0.3 dm^3 ) = 0.15 dm3 bar
= 0.15 dm^3 bar × 100 J /dm^3 bar
= 15.0 J