L) Q6) An isolated gas tank is divided into two equal partitions separated by an impermeable membrane. One side of the membrane is filled with gas, and the other side is vacuum. The membrane between the two partitions breaks. At the end of the process, the temperature and pressure are uniform throughout the gas tank. Calculate the final temperature, final pressure, and the change in entropy if the filled partition in the tank initially contains a. ideal gas at 32 MPa and 400 C, with a constant specific heat capacity of C;= 29.3 J/mol-K. [5] b. steam at 32 MPa and 400 C. [25] Question b: a) Starting off with the difference form of the energy balance: AU = Q+W; +WEC+Z'5MFHF We know that there is no work done to the system and that there is no mass flow or heat transfer. We also know that the system is isolated, this means that there is no change in internal energy. So the difference form of the energy balance can be reduced to: AU =0 We can use the following expression for the change in specific internal energy: AU = CpAT molK change in temperature, so AT = 0. We are given a value of C; = 29.3 and we know that AU = 0.This means that there would be no If there is no change in temperature, we know that the final temperature will also be equal to T;. T, = 400C We can calculate the final pressure using the ideal gas law. Since we know temperature remains the same: PI_VJ_ :P2V2 _)qu The question states that the volume is separated into 2 separate and equal portions. This means that when the gas diffuses fully into the vacuum section, the volume doubles: 10 V2 = 2VI Substituting the above relationship into Eq1: PV = P2 2V, P, =- 2 VI P1 P, = 2 32MPa P. = - 2 P2 = 16MPa To work out the change in entropy, I will assume that C is independent of temperature: AS = $ In P2 TI RIn 673 1.6 x 107 S(P,, V.) - S(P. V, ) = In - 8.314 In 673 3.2 x 107 45 = 5.76283 molk