Question 10 from the 11-year case study: a question on physical chemistry
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This post was last edited by forrestwong on 2016-7-16 at 18:02. In a tank with a volume of 0.35 m3, there is a vacuum pressure of 0.1 kPa; by quickly opening the valve on the tank, air from the atmosphere is drawn into the tank, until the pressure inside the tank equals the atmospheric pressure. During the inhalation process, the temperature inside the tank is always maintained at 25°C, the same as the atmospheric temperature. Using an air table (temperature-enthalpy diagram), the internal energy value of air at an initial state of 25°C and 0.1 kPa is 220 kJ/kg ； At the termination state of 25°C and 101.325 kPa, the enthalpy of air is 300 kJ/kg, and the molecular weight of air is 28.97. During this air intake process, what is the value of the heat released by the system to the environment in kJ? Hint: 1. The air inhalation process is an instantaneous dynamic process, which should be considered as an isothermal and isentropic gas process ； 2. The air quality inside the empty tank is considered to be zero. (A) 33 (B) 91 (C) 124 (D) 395 Solution: Analyze the gas by considering it as a system undergoing inhalation and expansion. First stage: isothermal and adiabatic, so Q1 = 0. Second stage (initial and final states are the same): △U = 0; therefore Q2 = W. △H2 = 0, so W = △(PV) = H – U = 300 – 220 = 80 KJ/kg. Given that V = 0.35 m3, P = 101.325, T = 298, and M = 28.97, we have PV = RTm/M, from which m = 0.4146 kg. Thus, Q = Q1 + Q2 = –W = –0.4146 × 80 = –33.168 KJ. How is W = △(PV) = H – U derived? ? ? Thank you
My approach to solving this problem is as follows: since it is an open system, the first law of thermodynamics for steady-flow systems applies, namely delta h = q + w. As there is no shaft work in this system, delta h = q = h2 – h1 = h2 – u1 = 80. Q = Q1 + Q2 = Q2 = m * q = pv/rt * M * q = 33.176 kJ
Actually, this question is a problem in chemical thermodynamics that needs to be solved using the generalized Gouy-Chapman equation! I don’t know if my picture was uploaded or not! It should be written quite clearly!
Can’t upload the image! In short, it’s d[m(U+u^2/2+gZ)]=Σ(H+u^2/2+gZ)δm+δQ+δW-pdV (I’m not sure if all these Greek letters can be displayed). This is not a steady-state flow situation, so differential equations are used. In the absence of shaft work, volume work, or changes in kinetic energy, the integral gives m(final)U(final)-m(initial)U(initial)=m(incoming)H(incoming)-m(leaving)H(leaving)+Q. According to the problem statement, m(initial)=0, m(leaving)=0, and m(final)=m(incoming); simplifying this yields mU(final)=mH(incoming)+Q. A negative value for Q indicates that the system releases heat!
Hey, hi. I can’t see it. Please resend the picture. Thank you.
Can’t send it! Just write out the process directly! Do you understand it now? All the other solutions do obtain an answer! But upon closer inspection, it’s all wrong!
How was this formula derived? Could you please write out the derivation process? I can’t understand it even at the first step. Thank you very much