Thermodynamics: Practice Problem & Solution
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
Solution Explained:
To solve this problem, we apply the core principles of Thermodynamics. Understanding the underlying formula is key to arriving at the correct answer below:
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: \(W = \frac{2R(300 - 600)}{5/3 - 1} = \frac{-600R}{2/3} = -900R\).
Leave a Reply