Ch 20: The First Law of ThermodynamicsWorksheetSee all chapters
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Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 04: Motion in Two and Three Dimensions
Ch 05: Friction, Inclines, Systems
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Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
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Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Concept #1: Work Done at a Constant Pressure

Transcript

Hey guys, in this video we're going to talk about the work done by gas in a piston at constant pressure. Let's get to it. Now consider an ideal gas at some initial pressure P stored in a cylinder with an area A as shown. So this area right here of the face of the cylinder is A and the gas happens to be in here at some pressure P. The gas therefore exerts a force on the piston to the left because the gas wants to expand. This is caused by those gas molecules colliding off of the wall so the forces to the left, this is the force due to the gas of just the pressure times the area that's by the definition of pressure. So this is PA. Now to stay in equilibrium the gas has to sorry the environment has to then exert a force back on to the piston head to keep it from moving, to keep it in equilibrium. If the gas's force was larger then the environmental force, the piston to move forward, the piston would expand in volume. If the environmental force was larger than the force by the gas the piston move inward and the cylinder's volume would decrease in equilibrium they're equal but in an opposite direction. So the environment must exert a force -PA. Now if the gas were to expand by some volume delta V right here I'm saying that the area is a constant so the change in volume has to be due to a change in length. This cylinder initially had some length L not, it ended with some length L final and that change in length the delta X is what produces that change in volume. So during this delta X the environment was exerting a force on the gas of -PA what we found here. So how much work is done by the environment on the gas that's how much work is done by the force due to the environment it's just -PA delta X just the force times delta X but negative because delta X is forward and the force is backwards. They point in opposite directions so the work is negative. Note that this area times delta X is just what the change in volume equals this just equals the area times delta X so the work done on the gas, very important, the work done on the gas by the environment at a constant pressure is -P delta V. This has the same sign convention that we use for work if this work is positive this is work done on the gas if this work is negative this is work done by the gas and that sign convention is very important because all of the conclusions that we draw are drawn from that sign convention. Remember guys like I said a positive work means that the gas gained energy. Negative work means the gas lost energy. The sign in the above question does actually agree with this if you were to go through and talk about what should happen to the gas's energy if the volume increases or if the volume decreases this work with the negative sign would exactly agree with our sign convention and you should always make sure that this is actually true. Let's do a quick example. 4 moles of an atomic hydrogen gas is compressed from 0.07 cubic meters to 0.01 cubic meters at a constant pressure of 2 times 10 to the 5 Pascals. How much work is done? Is the work done on or by the gas? At constant pressure the work is just -P delta V so the work is negative, the constant pressure is 2 times 10 to the 5, the change in volume is the final volume minus the initial and this equals 12000 joules or 12 kilojoules and it's positive. It's very important to recognize that this is positive because there is a decrease in the volume because this is negative that negative cancels with this negative so a decrease in the volume always going to be a positive work because the work is positive this means work was done on the gas. This is also very important to finish out the problem. Is the work done on or by the gas? It's done on the gas because by definition all positive works are work done on the gas. Alright guys that wraps up this discussion on work done at a constant pressure. Thanks for watching.

Concept #2: Thermal Processes of Ideal Gasses

Transcript

Hey guys, in this video we're going to introduce things called thermal processes. These are very popular questions asked in thermodynamics and they involve use of knowledge of work done at constant pressure and the First Law of Thermodynamics. Alright let's get to it. Now a thermal process is exactly what it sounds like. It's a process that a gas can undergo or any substance can undergo but we're typically gonna talk about ideal gases that can change the temperature. It doesn't have to change the temperature but it could change the temperature. Remember guys that the temperature of an ideal gas depends upon the pressure of the ideal gas and the volume of the ideal gas. This is just what the ideal gas equation says PV equals NRT. The temperature depends upon the pressure in the volume of the gas. When discussing thermal processes then it's convenient to plot pressure versus volume. This is known as a PV diagram and if we covered phase diagrams, if you covered them in your book, this is distinct from a phase diagram. A phase diagram was pressure versus temperature. This is pressure versus volume. Let me minimize myself so we can see right here. I have some hypothetical PV diagram on the right. The gas starts at an initial state V1P1 and the gas then goes at a constant pressure to increase its volume to V2. The pressure then increases at constant volume to V2P2 then the gas returns back to its original state. Remember guys that this kind of process that begins and ends at the initial state is a cyclic process. This is just an example of an arbitrary cyclic process, some hypothetical cyclic process as shown on a PV diagram, pressure versus volume. Now there are four types of five types sorry of thermal processes that you need to. No I was right the first time, four types. For some reason I thought this was five. There are four types of thermal processes that you guys need to know. There are isothermal processes, isobaric processes, isochoric processes and adiabatic processes. Iso which appears in the first three words means the same. Isothermal means constant temperature, isobaric means constant pressure maybe you heard of a barometer before, that baro that means pressure. Isochoric means constant volume and finally adiabatic means no heat transfer. So we could represent each of these as a statement. This would be delta T is zero, this would be delta P is zero, this would be delta V is zero and this would be Q is zero. Let's do a quick example oh by the way for cylic processes sorry, this I've talked about before, for cylic processes delta U is always zero. Remember that a cyclic process begins and ends in the same state because there is no change of state, there is no change in U, there is no change in the internal energy. The internal energy is a state function so it depends on the state of the gas. If you have no change in states you have no change in the state function. Let's do an example now. A gas undergoes a process as shown below, how much work is done? Is it work done on or by the gas? This is a two step process. We have an isobaric process first sorry isobaric and then we have an isochoric process, constant pressure first then constant volume. Isobaric processes mean that the work is just negative P delta V. The pressure is 1 times 10 to the 6, this is -1 times 10 to the 6. The change in volume is 0.05 to 0.03 so we end at 0.05 we begin at 0.03. So this is -20000 joules which is -20 kilojoules. Now step two is isochoric, if we look at this equation we can see that for no change in volume there's actually no work done. That was something that we had implicitly stated before several times that if there's no change in volume all the heat becomes a change in internal energy. So the work cannot be nonzero, the work has to be zero. For isochoric, the work is zero. This means that the total work done is just the work done in the isobaric process which is -20 kilojoules because no work was done after the isobaric process, during the isochoric process. Since the work was negative, this is work done by the gas the gas loses energy with a negative work. This is work done by the gas. Remember that that's how we finish the problem is it on or by the gas. Alright guys this wraps up the introduction into thermal processes. Thanks for watching.

Practice: 1 mole of a monoatomic ideal gas undergoes an isothermal expansion from 0.01 m3 to 0.05 m3 , requiring 300 J of heat to enter the gas. How much work is done? Is the work done by or on the gas? What is the change in the gas’ internal energy?

Practice: 3 moles of molecular hydrogen, H2, at 300 K undergoes an isochoric de-pressurization, from 5 x 105 Pa to 1 x 105 Pa. How much heat is exchanged during this process? Is the heat going into the gas or out of the gas? How much work is done during this process. Is the work done on or by the gas? How much does the internal energy of the gas change? Treat the molecular bond between the hydrogen atoms as elastic.