One mole of an ideal gas (with = 1.5 ) at an initial temperature T₀ expands isothermally until its volume doubles. It then expands adiabatically until its temperature drops to T₀ 3 . The ratio of the work done by the gas in the isothermal process to the work done in the adiabatic process is

Step-by-step solution

For the isothermal expansion of 1 mole of the gas at temperature T₀ to twice its initial volume, the work done is: W_ iso = n R T₀ ( V_f V_i ) = (1) R T₀ 2 Since the first process is isothermal, the temperature of the gas remains T₀ at the start of the adiabatic process. During the adiabatic expansion, the temperature drops from T₀ to T₀ 3 . The work done in the adiabatic process is: W_ adi = n R (T_i - T_f) - 1 W_ adi = (1) R (T₀ - T₀ 3 ) 1.5 - 1 W_ adi = R ( 2 T₀ 3 ) 0.5 = 4 3 R T₀ The ratio of the work done in t