n moles of a monoatomic ideal gas at an initial temperature T₀ is suddenly compressed in a thermally insulated container to 1 8 of its initial volume. The work done ON the gas during this process is: ( R is the universal gas constant)

Step-by-step solution

The process is sudden and in a thermally insulated container, so it is adiabatic. For an adiabatic process, the temperature and volume are related by: T_i V_i^ -1 = T_f V_f^ -1 For a monoatomic gas, = 5 3 , so - 1 = 2 3 . Given V_f = V_i 8 and T_i = T₀ , we can find the final temperature T_f : T_f = T₀ ( V_i V_f )^ -1 = T₀ (8)^ 2 3 Since 8 = 2^3 , we have (8)^ 2 3 = 2^2 = 4 . Thus, T_f = 4T₀ . The work done BY the gas in an adiabatic process is given by: W_ by = nR(T_i - T_f) - 1 W_ by = nR(T₀ - 4T₀) 5 3 - 1 = -3nR