Gases undergo different types of processes - cooling, heating, compression, or expansion, to name a few. They can undergo these processes under different conditions as well – under constant pressure, constant temperature, constant volume, no heat addition or removal (adibatic) etc. In this problem, you are given an ideal gas that is tasked to undergo a reversible cyclic process (please see the slides on what is reversibility; note that cyclic processes return to its initial condition). You will solve the Q (heat), W (work), AU (change in internal energy), and AH (change in enthalpy) for each steps in the reversible cycle and for the reversible cycle in total. Additionally, what if the entire cycle is irreversible? You will also be asked to solve the Q, w, AU, AH for each steps in the irreversible cycle and for the irreversible cycle in total. You will learn that some of the values remain the same whether it is reversible or irreversible and those are values of the state functions. An ideal gas, initially at 30 deg C and 100 kPa, undergoes the following cyclic processes in a closed system: a. In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a constant pressure of 500 kPa to 30 deg C, and finally expanded isothermally to its original state. Calculate the Q, W, AU, and AH for each step of the process and for the cycle with all answers in kl/mol. Take Cp = (7/2)R and Cv = (5/2)R. Hint: Better to draw a box diagram to visualize the states and the steps involved in the cyclic process. As a cycle, it should return to its initial condition after all the steps. Solve first for each step of the reversible cycle (e.g. Step 1: Adiabatic compression from 100 kPa to 500 kPa). Then for this step, enumerate the required Q, W, AU, AH and their equivalent equation based on the condition. For e.g. for adiabatic compression: W = dU; Q = 0; dU = Cv dT; dH = Cp dT Do similarly for all the steps. Note that you can find the applicable equations for the other conditions in the slides provided in the LMS. Once you have computed for each steps, you can make a tabular version for the overall cycle. The overall Q is the sum of all the individual Q. The overall W is the sum of all the individual W and so on. Note that since this is a cycle, the overall CHANGE in H and U should be 0. Finally, the overall CHANGE in U = overall Q + overall W. b. The cycle traverses exactly the same changes of state, but each step is irreversible with an efficiency of 80% compared with the corresponding mechanically reversible process. Note: The initial step can no longer be adiabatic. Calculate the Q, W, AU, and AH for each step of the process and for the cycle with all answers in kJ/mol. Take Cp = (7/2)R and Cv = (5/2)R.
Gases undergo different types of processes - cooling, heating, compression, or expansion, to name a few. They can undergo these processes under different conditions as well – under constant pressure, constant temperature, constant volume, no heat addition or removal (adibatic) etc. In this problem, you are given an ideal gas that is tasked to undergo a reversible cyclic process (please see the slides on what is reversibility; note that cyclic processes return to its initial condition). You will solve the Q (heat), W (work), AU (change in internal energy), and AH (change in enthalpy) for each steps in the reversible cycle and for the reversible cycle in total. Additionally, what if the entire cycle is irreversible? You will also be asked to solve the Q, w, AU, AH for each steps in the irreversible cycle and for the irreversible cycle in total. You will learn that some of the values remain the same whether it is reversible or irreversible and those are values of the state functions. An ideal gas, initially at 30 deg C and 100 kPa, undergoes the following cyclic processes in a closed system: a. In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a constant pressure of 500 kPa to 30 deg C, and finally expanded isothermally to its original state. Calculate the Q, W, AU, and AH for each step of the process and for the cycle with all answers in kl/mol. Take Cp = (7/2)R and Cv = (5/2)R. Hint: Better to draw a box diagram to visualize the states and the steps involved in the cyclic process. As a cycle, it should return to its initial condition after all the steps. Solve first for each step of the reversible cycle (e.g. Step 1: Adiabatic compression from 100 kPa to 500 kPa). Then for this step, enumerate the required Q, W, AU, AH and their equivalent equation based on the condition. For e.g. for adiabatic compression: W = dU; Q = 0; dU = Cv dT; dH = Cp dT Do similarly for all the steps. Note that you can find the applicable equations for the other conditions in the slides provided in the LMS. Once you have computed for each steps, you can make a tabular version for the overall cycle. The overall Q is the sum of all the individual Q. The overall W is the sum of all the individual W and so on. Note that since this is a cycle, the overall CHANGE in H and U should be 0. Finally, the overall CHANGE in U = overall Q + overall W. b. The cycle traverses exactly the same changes of state, but each step is irreversible with an efficiency of 80% compared with the corresponding mechanically reversible process. Note: The initial step can no longer be adiabatic. Calculate the Q, W, AU, and AH for each step of the process and for the cycle with all answers in kJ/mol. Take Cp = (7/2)R and Cv = (5/2)R.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Transcribed Image Text:Gases undergo different types of processes – cooling, heating, compression, or expansion, to name a few. They can undergo
these processes under different conditions as well – under constant pressure, constant temperature, constant volume, no
heat addition or removal (adibatic) etc.
In this problem, you are given an ideal gas that is tasked to undergo a reversible cyclic process (please see the slides on what
is reversibility; note that cyclic processes return to its initial condition). You will solve the Q (heat), W (work), AU (change in
internal energy), and AH (change in enthalpy) for each steps in the reversible cycle and for the reversible cycle in total.
Additionally, what if the entire cycle is irreversible? You will also be asked to solve the Q, w, AU, AH for each steps in the
irreversible cycle and for the irreversible cycle in total. You will learn that some of the values remain the same whether it is
reversible or irreversible and those are values of the state functions.
An ideal gas, initially at 30 deg C and 100 kPa, undergoes the following cyclic processes in a closed system:
a. In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a constant pressure
of 500 kPa to 30 deg C, and finally expanded isothermally to its original state. Calculate the Q, W, AU, and AH for each step
of the process and for the cycle with all answers in kJ/mol. Take Cp = (7/2)R and Cv = (5/2)R.
Hìnt: Better to draw a box diagram to visualize the states and the steps involved in the cyclic process. As a cycle, it
should return to its initial condition after all the steps. Solve first for each step of the reversible cycle (e.g. Step 1:
Adiabatic compression from 100 kPa to 500 kPa). Then for this step, enumerate the required Q, W, AU, AH and
their equivalent equation based on the condition. For e.g. for adiabatic compression:
W = dU; Q = 0; dU = Cv dT; dH = Cp dT
Do similarly for all the steps. Note that you can find the applicable equations for the other conditions in the slides
provided in the LMS. Once you have computed for each steps, you can make a tabular version for the overall cycle.
The overall Q is the sum of all the individual Q. The overall W is the sum of all the individual W and so on. Note
that since this is a cycle, the overall CHANGE in H and U should be 0. Finally, the overall CHANGE in U = overall Q +
overall W.
b. The cycle traverses exactly the same changes of state, but each step is irreversible with an efficiency of 80% compared
with the corresponding mechanically reversible process. Note: The initial step can no longer be adiabatic. Calculate the Q,
W, AU, and AH for each step of the process and for the cycle with all answers in kJ/mol. Take Cp = (7/2)R and Cv = (5/2)R.
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