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 1st law of thermodynamics and the ideal gas equation .
Thanks
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
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