One mole of an ideal gas at 300 ~K is expanded isothermally from an initial pressure of 10 ~atm to a final pressure of 1 ~atm via two different paths: Path A: Reversible expansion. Path B: Irreversible expansion against a constant external pressure of 1 ~atm . The magnitude of the difference in work done between the two paths, |w_ rev | - |w_ irr | , is: Given: R = 8.3 ~J K⁻¹ mol⁻¹ and 10 = 2.3 .

Step-by-step solution

For Path A (reversible isothermal expansion), the work done is: w_ rev = -nRT ( P_i P_f ) w_ rev = -1 8.3 300 ( 10 1 ) w_ rev = -2490 2.3 = -5727 ~J For Path B (irreversible isothermal expansion against constant external pressure), the work done is: w_ irr = -P_ ext (V_f - V_i) Using the ideal gas law, V = nRT P , we can express the volumes in terms of pressures: w_ irr = -P_ ext ( nRT P_f - nRT P_i ) = -P_ ext nRT ( 1 P_f - 1 P_i ) Substitute the values ( P_ ext = 1 ~atm , P_f = 1 ~atm , P_i = 10 ~atm ): w_ irr =