Question

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. 

Answer 1

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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