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Physics Chapter 15
| Question | Answer |
|---|---|
| Zeroth Law of Thermodynamics | Two systems individually in thermal equilibrium with a third system are also in thermal equilibrium with each other. |
| First Law of Thermodynamics | The change in internal energy (delta U) of a closed system is due to the heat added or removed (Q) and/or work done on or by the system (W). |
| Change in internal energy equation | Delta U = Uf - Ui = Q - W |
| When is Q positive | Heat flows into the system Examples: Heating a gas Putting a cold object in hot water Endothermic process |
| When is Q negative | Heat flows out of the system Examples: Cooling a gas Hot object placed in cold surroundings Exothermic reaction |
| When is W positive | System does work on surroundings Examples: Gas expands against a piston System pushes something outward |
| When is W negative | Surroundings do work on the system Examples: Gas compressed Piston pushed inward |
| Tip for work and volume | Anytime the volume is constant, the work isn't going to change either. |
| Equations for Cp and Cv | In thermodynamic equations, we tend to substitute mass for moles during Q problems. Q = nCvdelta T at constant volume and Q = nCpdelta T at constant pressure. Cv = 3/2R and Cp = 5/2R |
| Isothermal Process | The temperature of the gas remains constant. For example, when a container is surrounded by hot sand, the temp of the gas remains constant. Change in U = 0 Q = W W = equation given in sheet |
| Isobaric Process | The pressure is constant; ex: cylindrical container with movable piston. You put a weight on the piston, meaning that P = F/A, and since force comes from the weight of the object, all variables remain constant. |
| Isobaric Process Equations | Change in U = nCvdelta T or Q - W Work = PdeltaV Q = nCpdelta T PV diagram = horizontal line, allowing you to easily take the area and find work |
| Isochoric Process | Volume remains constant. W = 0 Q = nCvdelta T change in U = nCvdelta T PV diagram = straight vertical line with no curve |
| Tip about Rigid Containers | Anytime you hear the word rigid, you know the volume is constant and the process is iscohoric! |
| Adibatic Process | No heat flows into or out of the system. Q = 0 W = -delta U Delta U = -W Both equations on sheet are given. PV Graph = steeper than isothermal Gamma = 5/3 |
| Second Law of Thermodynamics | 1) heat flows spontaneously from a hot object to a cold object but not vice versa. 2) it is impossible to construct a heat engine, which operates in a cycle, that is 100%. Some input heat will be turned to useful work but the rest will be waste Q. |
| Efficiency Equations | Qh = W + Qc e = W/Qh = 1 - Qc/Qh Qh = input heat (from the top) W = work (to the side) Qc = waste heat (goes down) |
| Tip about spontaneous reactions | Heat can only move from a hot reservoir to a cold one spontaneously, not vice versa. |
| Carnot Engine Rule | In order for the Carnot engine to be 100% efficient, the temperature of the exhaust heat needs to absolute zero -> not possible. |
| 1st Process (Carnot Engine) | Isothermal, engine acquires energy Qh at the temperature Th. |
| 2nd Process (Carnot Engine) | Temperature is reduced from Th to Tc adibatically with no heat exchanged with the environment. |
| 3rd Process (Carnot Engine) | Isothermal, Qc is rejected to a cold reservoir. Maintained at temp Tc. |
| 4th Process (Carnot Engine) | System adibatically returns from Tc to Th. |
| Third Law of Thermodynamics | It is IMPOSSIBLE to reach absolute zero in a finite number of steps, as each step to lower the temperature would require an infinite number of additional steps to get closer to the unattainable zero point. |
| Refigerators | Operates by opposite means -> removes heat from a low temperature (cold) reservoir and exhausts the heat to a high temperature (hot) reservoir. Work has to be done to make the heat flow in an opp direction. |
| Useful coefficient of performance equations | n = Qc/W or n = Qc/(Qh - Qc) n ideal = Tc/(Th - Tc) Qc/Tc = Qh/Th |
| Entrophy | Quantitive measure of the disorder in a system. Proportional to heat added to the system and inversely proportional to temp added to the system. Delta S = Q/T. |
| Change in Entrophy in Hot Temp | Delta Sh = -Qh/Th |
| Change in Entrophy in Cold Temp | Delta Sc = +Qc/Tc |
| Law of Thermodynamics Concerning Entrophy | The entrophy of an isolated system never decreases. It can only stay the same or increase. Delta S = delta Ssys + delta Senv > 0 |
| Easy way to find entropy change of the universe | Add entropy change of Sh and Sc together |
| Helpful Tip for Constant Volume Problems | You can use the equation Q = 3/2V delta P |
| Helpful Tip for Constant Pressure Problems | You can use the equation Q = 5/2P delta V |
| To find how much heat has been added to a gas... | You can use Q = W + delta U |
| How you do convert m^3 to L? | You multiply by 1000 |
| Tip for Isochoric Processes | Delta U = Q |
| For questions like number 10, what is the trick? | You can replace W with PV like in the PV = nRT equation |
Created by:
smurtab