This preview shows page 1. Sign up to view the full content.
Unformatted text preview: figures, we obtain: Δ T = 1.14 K 4 Question 11: (10 points) 2.00 mol of a monoatomic ideal gas (CVm = 3/2 R) undergo an irreversible adiabatic expansion under 5
constant external pressure from an initial state given by P1 = 1.0 x 10 Pa and T1 = 240.5 K until the final 4
equilibrium state where the pressure is P2 = 1.0 x 10 Pa. Part a. Show that the final temperature T2 is equal to 154K (5 points). For an adiabatic expansion at constant external pressure, the work is equal to the change in energy, which for an ideal gas is dU = nCVm dT. Hence,  pext dV = nCVm dT This equation is integrated as:  pext (V2 – V1) = nCVm (T2 – T1) Expressing the volumes in terms of the temperature and pressure, we get:  pext (nRT2 / p2 – nRT1 / p1) = n(3/2)R (T2 – T1) which is simplified as:  pext (T2 / p2 – T1 / p1) = (3/2)(T2 – T1) and noting that pext = p2 and p1 = 10 p2  T2 + T1 / 10 = (3/2)T2 – (3/2)T1 Hence, (5/2) T2 = (3/2)T1 + T1 / 10 and T2 = 153.92 K Part b. Calculate the change in entropy for the gas, for the surroundings and show that the process is spontaneous. The change in entropy must be calculated for a reversible path starting at P1, T1, V1 and ending at P2, V2, T2. 5
3
V1 = nRT1/P1 = 2.00 mol x 8.3145 J/mol.K x 240.5 K / 1.0x10 Pa = 0.039993 m 4
3
V2 = nRT2/P2 = 2.00 mol x 8.3145 J/mol.K x 153.92 K / 1.0x10 Pa = 0.255954 m The reversible path could consist of...
View
Full
Document
 Spring '07
 AREsker
 Physical chemistry, pH

Click to edit the document details