Question

Calculate the maximum work that could be performed by 2 moles of an ideal gas undergoing a reversible adiabatic expansion from 10³ cm to 100³cm.

Must combine the 1^st law of thermodynamics and the ideal gas equation .

Thanks

Answer 1

Given data

Number of moles of gas (n) = 2 moles

initial volume V₁= 103 cm³

final volume v₂ = 1003

ideal gas equation is given as PV=nRT

As temprature is not given T= 298 k (standard)

Gas constant R = 8.314 J/K

According to firs law of thermodynamics

Internal enegy change U = heat supplied to the system (q) + work done on the system (W)

in adiabatic process q = 0 =zero

Internal enegy change = 0 + w

now,

work done is given as W = P x (V₂ - V₁)

for adiabatic reversible condition It need to be integrated

taking PV= nRT

P=(n)RT/V

and it is Maximum Work given as

Wmax =-

W max = -2.303 nRT logV₂/V₁

          = -2.303*2*8.314*log 1003/103

          = -2.303*2*8.314*log 9.738

          = -2.303*2*8.314* 0.9885

W max = -37.854 J

If temprature is given put its value and calculate