A mixture of helium and neon gas is expanded from a volume of 47.0 L to a volume of 93.0 L, while the pressure is heid constant at 97.0 atm. Calculate the work done on the gas mixture.

W = - 452.112 kJ

Explanation:

given data

initial volume v1 = 47 L

final volume v2 = 93 L

external pressure = 97 atm

solution

we get here work done that is express as

W = - P (V2 - V1)           1

put here value and we get

as V2 - V1 = 93 L - 47 L = 46 L

1 atm = 101325 pascal

41 atm = 9828525 pascal

46 L = 0.046 m³

W = - 9828525 pascal × 0.046 m³ = 452112.15 J       ( 1KJ = 1000 J )

so

W = - 452.112 kJ

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"