Solution:
Work done in an adiabatic process is \(W = \frac{nR(T_1 - T_2)}{\gamma - 1}\). For monoatomic gas, \(\gamma = 5/3\). Substituting the parameters: \(W = \frac{2R(300 - 600)}{5/3 - 1} = \frac{-600R}{2/3} = -900R\).
Two moles of an ideal monoatomic gas undergoes an adiabatic process from temperature \(300\text{ K}\) to \(600\text{ K}\). Work done by this ideal gas in the process is
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