Suppose 3.00 moles of a monatomic ideal gas expand adiabatically, and its temperature decreases from 397 to 258 K. Determine (a) the work done (including the algebraic sign) by the gas, and (b) the change in its internal energy.
In Adiabatic process work-done is given by:
W = n*R*dT/(1 - y)
n = number of moles = 3.00
R = gas constant = 8.314
dT = T2 - T1 = 258 - 397 = -139 K
for a monatomic gas, y = 1..67
So,
W = 3*8.314*(-139)/(1 - 1.67)
Work-done, W = 5174.5 J
Part B.
Using first law of thermodynamics
dQ = dU + W
Since in adiabatic process, dQ = 0, So
dU + W = 0
dU = -W
Change in internal energy, dU = -5174.5 J
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