Question

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.

Answer 1

W = - 452.112 kJ

Step-by-step 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