Adiabatic expansion of an ideal gas A. For certain ideal gas Cp=8.58 cal/mol-deg. What will be the final temperature and volume when two moles of the gas at 20 degrees celsius and 15 atmosphere are allowed to expand adiabatically and reversibly to 5 atm pressure? Find W, ∆U, and ∆H.
B. Suppose that the expansion takes place adiabatically against a constant pressure of five atm, what will be the final volume and temperature of the given gas?
I already solved part A but I have no clue for part B. Especially on how to get the final temperature and pressure.
In the following answer I am using the Physics convention of work done. Where, $W= \int PdV$ and the first law of thermodynamics is as follows: $q = \Delta U + W$. If you want answers in the chemistry conventions than just reverse the sign of the work done. Internal energy change and heat change will be same in both conventions.
As you said you have done the first part, I will continue with the second.
First of all, $C_V=C_P - R = 8.58- 2 = 6.58$
As the process is adiabatic,
$$q=0\\
\Delta U=-W\\
nC_V(T_2-T_1)=-nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
T_2= 247.67 K = -25. 53 ^\circ C\\\\$$
$\\Now, \\\\$
$$W= P_{ext} \Delta V\\
W= nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
W= 599.2 cal\\\\$$
$$\Delta U= -W \\
\Delta U= -599. 2 cal\\\\$$
$$\Delta H= \Delta U + \Delta (PV) \\
\Delta H= \Delta U + \Delta (nRT)\\ 
\Delta H= \Delta U + nR\Delta T\\
\Delta H= (-599. 2 -181. 32) cal\\
\Delta H= -780. 52 cal\\\\$$