Question

A piston containing 0.75 moles of a perfect gas at 25 degrees C is moved isothermally from a starting volume of 8.305 L.

Calculate the work in joules if the piston is moved irreversibly against a constant external pressure of 4.100 atm. We weren't giving final volume so I am unsure of what to do. I tried to use P1V1 = P2V2 because the final pressure is equal to external pressure in an irreversible process. But then my final volume decreases when it's supposed to be expanding.

Use this information to answer the following: A piston containing 0.75 moles of a perfect gas at 25.00 °C is moved isothermally from a starting volume of 8.305 L. a) Calculate the work (w) in joules if the piston is moved irreversibly against a constant external pressure of 4.100 atm. (Note that the system will come to equilibrium at a final volume when the external pressure is equal to the internal pressure). b) Calculate the work (w), heat transfer (q), change in internal energy (AU), and change in enthalpy (AH) in joules for the process above if the movement is done reversibly instead of irreversibly. All other variables are the same.

Answer 1

Giver, n=0.75 mol T= 25°c = 298k V = 8.305L "PV=nRT & P = ART = 0.75*0.0821 x 298 mol Katm motky k 8.305k P = 2.21 atm P2 = Pext = 4. 1 atm Since process is isothermal Piv, = PzV₂ = ART Constant V2 = Pivi = 2.212fm, x8.3052 P2 4. 1 atom [V₂ = 4.47527 in final pressure is more than ste external pressure compression will take place Volume final decreases . (a) Wire =-Pext (V2 - Vy) --4.1 (4.475 - 28-305) L atm

= +15. 70 115 Latm (:1 Latm= 101.3255) [Wise + 1590.925] Compression taking place Work is being done on the system W= tve # ferocess is Reversible Wrew = - ART ln (12) = -0.75x8.314x29814.475 molol Jelmat K (8.305 | Wrew=+114qJ] Using 14 law of thermodynamics, AU=q+W suančudt (.dT=0 iso the smal process) AU=0 q=- w = q = -11495) Work is done on the system and as a result of which heat is released by the system. TH= nCp dT (:&T=0) AH=0 Theis, the values of enthalpy charge and internal energy change will be zero since process is iso thermal.