# Reversible & Irreversible Processes

## Presentation on theme: "Reversible & Irreversible Processes"— Presentation transcript:

Reversible & Irreversible Processes
Contributions by: John L. Falconer Department of Chemical and Biological Engineering University of Colorado Boulder, CO Supported by the National Science Foundation

When the red stops are removed, the ideal gases expand, and the pistons move until they hit the black stoppers. Each system is adiabatic. Which system has the most lost work? A B C 2 kg Vacuum Gas 1 kg Vacuum Gas Vacuum Gas A B C Same for all systems ANSWER: C. The system with no weight on the piston. This system has the greatest change in entropy, and the greatest increase in entropy makes that system the most irreversible, which translates into the most lost work.

In these piston-cylinder systems, when the red stop is removed, the ideal gas expands, and the piston moves until it hits the black stopper. Each system is adiabatic and starts at 10 atm and 25°C. Which has the highest final temperature? A B C 2 kg Vacuum Gas Piston Block 1 kg Vacuum Gas Vacuum Gas A B C All the same ANSWER: C. The system with no weight on the piston. The least work is done in this system so there is the lowest temperature drop.

A pile of sand sits on a piston and the weight of the sand keeps the gas compressed in the cylinder. Which method to remove the sand expands the gas most reversibly? Gas Piston Sand All at once. One grain at a time. ¼th of the sand per minute. Same for all methods ANSWER: B. One grain at a time. This method is the slowest and thus closest to reversible because it minimizes pressure gradients.

An ideal gas expands from 2 bar to 1 bar adiabatically but irreversibly. Compared to an adiabatic, reversible process from 2 bar to 1 bar, the final temperature will be _____________ for the irreversible process. higher lower the same ANSWER: B. higher for the irreversible process. Because the irreversible process does less work and has more “lost work,” there is less of a decrease in internal energy.

All regions can be reached
The curve represents an adiabatic reversible process for an ideal gas. For any starting point on the line, which regions cannot be reached by an adiabatic irreversible process? U V ABOVE U V U V BELOW A B C All regions can be reached ANSWER: B. The graph of the area below the curve. For a given change, the irreversible process will have more internal energy (with both compression or expansion), so the irreversible process will always be within the area above the reversible line.

Which of these processes could run under continuous, steady-state operation?
Q W Q 1 2 & 3 3 4 1 & 4 #1 #2 W Q W Q ANSWER: D. 4. Work can be converted completely into heat. 1 and 2 violate 1st law, 3 violates 2nd. #3 #4

#1 W QH hot cold QC W Q #2 W Q W Q #3 #4
Which of the following processes could run under continuous, steady-state operation? #1 W QH hot cold QC W Q 1 2 & 3 3 4 1 & 4 #2 W Q W Q ANSWER: D. 4. Work can be converted completely into heat. 1 and 2 violate 1st law, 3 violates 2nd. #3 #4

W QH hot cold QC #1 W QH hot cold QC #2 W QH hot cold QC W QH hot cold
Which of the following processes could run under continuous, steady-state operation? W QH hot cold QC #1 W QH hot cold QC 1 & 2 3 & 4 3 4 1 & 3 #2 W QH hot cold QC W QH hot cold QC ANSWER: B. 3 & 4. Processes #3 and #4 are heat pumps and heat engines, but 1 and 2 violate first law. #3 #4

W QH hot cold QC #1 W QH hot cold QC #2 W QH hot cold QC W QH hot cold
Which of the following processes could run under continuous, steady-state operation? W QH hot cold QC #1 W QH hot cold QC 1 3 1 & 3 4 1 & 2 #2 W QH hot cold QC W QH hot cold QC ANSWER: C. 1 & 3. Process #1 is a heat engine, and for process #3, work can be converted completely into heat. Process 4 violates 2nd law (all heat converted to work), process 2 violates 2nd law (a heat pump that generates work/electricity). #3 #4

A gas goes from state A to state B in a reversible adiabatic process
A gas goes from state A to state B in a reversible adiabatic process. It then goes from B back to A by a different pathway that is irreversible and not adiabatic. The entropy change for the gas for the irreversible pathway is _____________ zero. V P A B greater than less than equal to Irrev. ANSWER: C. equal to zero. Entropy is a state function, so the entropy change for one process is the negative of that for the other. The entropy change for the reversible adiabatic process is zero. Rev.

A gas goes from state A to state B in a reversible adiabatic process
A gas goes from state A to state B in a reversible adiabatic process. It then goes from B back to A by a different pathway that is irreversible and adiabatic so that the entropy change of the gas is zero. Is this possible? V P A B Yes No Cannot tell Irrev. ANSWER: A. Yes. Entropy is a state function, so the entropy change for one process is the negative of that for the other. The entropy change for the reversible adiabatic process is zero, so it must be zero for the irreversible and adiabatic process to return to state A. Rev.

Two reversible pathways are shown.
Which one has the larger value of Q? 1  2  3  1 1  4  5  1 Both have the same Q ANSWER: B. 1  4  5  1. More work is done in process B (greater PV area), so more heat must be added since the change in internal energy is zero for a cycle.

Two reversible pathways are shown.
Which one has the larger value of Q? 1 2 3 4 5 P V 1  2  3  1 1  4  5  1 Both have the same Q ANSWER: A. 1  2  3  1. The Q terms are negative for both processes, but Q is more negative for B (greater PV area) because more heat must be removed since the change in internal energy is zero for a cycle. The cycle for B has a greater absolute value of Q.

Adiabatic expansions and compressions are shown for an ideal gas
Adiabatic expansions and compressions are shown for an ideal gas. One curve is reversible and one is irreversible in each figure. Which ones are the irreversible curves? 1 2 P T start 1 & 3 1 & 4 2 & 3 2 & 4 3 4 P T start ANSWER: D. 2 & 4. For both processes, the irreversible process always runs at the higher temperature. For compression, it takes more work to compress irreversibly and that work is transferred into internal energy. For expansion, less work is generated so there is a less negative change in internal energy (and thus smaller decrease in temperature).

start P V 2 1 3 The three curves represent isothermal, reversible adiabatic, and irreversible adiabatic processes for a gas. Which curve represents the irreversible adiabatic process? 1 2 3 Cannot be determined ANSWER: A. Curve 1. Curve 3 is isothermal (the smallest volume for a given pressure because the final temperature is lowest since heat has to be removed). Curve 1 is irreversible adiabatic because temperature is greater (the volume is less at a given pressure) than for reversible adiabatic (which then must be curve 2).

start P V 1 3 2 The three curves represent isothermal, reversible adiabatic, and irreversible adiabatic processes for a gas. Which curve represents the irreversible adiabatic process? 1 2 3 Cannot be determined ANSWER: B. 2. Curve 1 is isothermal (the greatest pressure for a given volume because the final temperature is greatest since heat must be added). Temperature drops for the adiabatic processes because energy is removed from system as work. Curve 2 is irreversible adiabatic because temperature does not drop as much for irreversible adiabatic (curve 3 is reversible adiabatic)

An ideal gas goes from 10 atm and 50°C to 1 atm and 100°C by two pathways. Path A is reversible, it is a constant volume step followed by a constant pressure step. Path B is irreversible, and is a constant pressure step followed by a constant volume step. Which pathway has the larger entropy change for the gas? V P 10 atm 50°C 1 atm 100°C A A B They are the same ANSWER: C. They are the same. Entropy is a state function. P V 10 atm 50°C 1 atm 100°C B

An inventor claims he can convert 1000 J of heat into 1000 J of work
An inventor claims he can convert 1000 J of heat into 1000 J of work. Does this violate the second law? Yes Yes, if he claims to do it continuously No Cannot tell ANSWER: B. Yes, if he claims to do it continuously.

Slowly increasing pressure on the gas phase reaction 2A(g) 2B(g)
Which process is the most irreversible? Slowly increasing pressure on the gas phase reaction 2A(g) B(g) Melting ice at 0°C by removing heat. Slowly blowing up a balloon. Slowly mixing 40°C water with 42°C water. Boiling water in a piston/cylinder at 1 atm, 100°C. ANSWER: D. Slowly mixing water. Cannot un-mix the water easily and reverse the process.

Which process is closest to being reversible?
A. Slowly dissolving NaCl in water B. Slowly burning carbon in O2 at 300°C C. Slowly compressing a gas in a piston-cylinder so that the gas heats up D. Expanding a gas through a nozzle to lower pressure ANSWER: C. Slowly compressing a gas in a piston-cylinder so that the gas heats up. This is the only process that could be made to go in the opposite direction by a slight change in driving force.

An ideal gas is compressed adiabatically and irreversibly
An ideal gas is compressed adiabatically and irreversibly. The final temperature of this process relative to that for an adiabatic reversible process to the same final pressure is ____________. higher lower the same ANSWER: A. higher. The temperature is higher for the irreversible process because more work has to be added to get to the same pressure and the work transfers into internal energy.

The work for irreversible compression of an ideal gas in a piston-cylinder system is _______ the work for reversible compression. higher than lower than the same as ANSWER: