An ideal monoatomic gas undergoes an isothermal expansion at an initial temperature of 127^ C and does W joule of work. The same gas then undergoes an adiabatic expansion from the same initial temperature, doing the exact same amount of work W . If the final temperature of the gas after the adiabatic expansion is -33^ C , the ratio of the final volume to the initial volume for the isothermal expansion is:

Step-by-step solution

Convert the given temperatures into Kelvin: T_i = 127 + 273 = 400 K T_f = -33 + 273 = 240 K For a monoatomic ideal gas, the molar heat capacity at constant volume is C_V = 3 2 R = 1.5R . The work done during the adiabatic expansion is: W_ adia = nC_V(T_i - T_f) = n(1.5R)(400 - 240) = n(1.5R)(160) = 240nR The work done during the isothermal expansion is: W_ iso = nRT_i ( V_f V_i ) = nR(400) ( V_f V_i ) Given that the work done in both processes is the same, we equate the two expressions: 400nR ( V_f V_i ) = 240nR Di