One mole of an ideal monoatomic gas at an initial temperature T₀ undergoes an adiabatic expansion. If the work done by the gas during this process is 3 4 RT₀ , what is the ratio of its final volume to its initial volume?

Step-by-step solution

For an ideal monoatomic gas, the adiabatic index is = 5 3 . The work done in an adiabatic process is given by: W = nR(T_i - T_f) - 1 Given n = 1 , T_i = T₀ , and W = 3 4 RT₀ , we can substitute these values: 3 4 RT₀ = 1 R(T₀ - T_f) 5 3 - 1 3 4 RT₀ = 3 2 R(T₀ - T_f) Dividing both sides by 3 2 R : T₀ 2 = T₀ - T_f T_f = T₀ 2 For an adiabatic process, the temperature and volume are related by TV^ - 1 = constant . T₀ V₀^ - 1 = T_f V_f^ - 1 Substitute T_f = T₀ 2 and - 1 = 2 3 : T₀ V₀^ 2/3 = T₀ 2 V_f^ 2/3 ( V_f V₀ )^ 2/3