The First Law of Thermodynamics

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You give a gas five hundred joules of heat. So its internal energy goes up by five hundred joules, right? Well, that is not always true. In one problem on this page, it goes up by only three hundred. So where did the other two hundred joules go? Did energy just vanish? One simple line answers that: the first law of thermodynamics. Questions on it come up often in JEE, so let's begin.

0:31The first law: energy is conserved

So, picture a gas in a cylinder with a piston. We give it some heat. Capital Q is the heat given to the system, okay? Now think. Where can this energy go? It can go to two places. Some of it stays inside and changes U, the internal energy. Its change is delta U; delta, the little triangle, means change in. The rest leaves as work, capital W, done by the system when the gas pushes the piston out. Clear? So, Q equals delta U plus W, okay? This is the first law of thermodynamics. It is just conservation of energy: the heat that comes in is either stored inside or spent as work. With me? Now, our puzzle. The gas absorbed five hundred joules and did two hundred joules of work; capital J is the joule. The double arrow means so: delta U equals Q minus W, five hundred minus two hundred, three hundred joules. The missing two hundred joules left as work. Energy never vanished! For a tiny part of a process, we write d Q equals d U plus p d V, okay? The small d means a tiny change. Small p is the pressure and d V a tiny change in volume, so p d V is the tiny bit of work done by the gas. Got it?

1:50Signs: the classic trap

Now, the most important thing here: signs. Q is the heat given to the system, so heat absorbed is positive, Q greater than zero, and heat released is negative, Q less than zero. Okay? And W is the work done by the system. So if we push the piston in and do work on the gas, what is the sign of W? Negative, exactly! W is positive when the gas expands, and negative when the gas is compressed. Students often write plus here and lose easy marks, so be careful with this one. Right? Now, chemists use the work done on the system instead: W with the small word on written below it. Then the law reads delta U equals Q plus W on. Looks different, doesn't it? But it is the same law. Work done on the gas is minus the work done by it: W on equals minus W, okay? Put that in, and you get back our law. Both forms say energy is conserved. In this video we use Q equals delta U plus W, with W done by the system. Clear?

2:59ΔU depends only on the end states

Next, a big idea. Internal energy is a state function: it depends only on the state of the system, not on how it got there. For an ideal gas, U depends on the temperature, capital T, alone. Okay? So between the same two states, delta U is the same for every path. With me? Now, what about Q and W? Are they also the same for every path? No, they are not! They depend on the path. Only their difference, Q minus W, is fixed, because that difference is delta U itself. Got it? Now a special case: a cycle, where the system comes back to its starting state. For one full cycle, what is delta U? It is zero, exactly! Same start, same end, so no change in internal energy. Right? Put delta U equal to zero in the first law, and Q equals W, okay? So in a cycle, the net heat absorbed equals the net work done. Clear? For example, a gas absorbs nine hundred joules in one part of a cycle and releases six hundred in another. Net heat, three hundred joules, so the net work done is three hundred joules. It is really that simple!

4:14Example 1: work done on gas, heat released, find delta U

Now, example one. Three hundred joules of work is done on a gas, and it releases one hundred joules of heat. Find delta U, the change in internal energy. Watch the signs carefully. Pause the video and try it first. Done? Okay, let's check. Many students write: Q is one hundred, W is three hundred, so delta U is one hundred minus three hundred, minus two hundred joules. Is that right? No, no, careful here! Both signs are wrong. The gas releases heat, so Q is minus one hundred joules. Okay? And work is done on the gas, so the work done by the gas is negative: W is minus three hundred joules. With me? Now the first law: delta U equals Q minus W, so minus one hundred, minus minus three hundred, that is plus two hundred joules. It makes sense, right? Three hundred joules came in as work, and only one hundred left as heat.

5:14Example 2: same delta U on two paths, heat on the way back

Now a harder one, example two. This time, a system goes from state i, the initial state, to state f, the final state, along path i a f, absorbing two hundred joules and doing eighty joules of work. Along path i b f, it does thirty joules of work. U i, U with a small i written below, is the internal energy at the start: fifty joules. Find U f, and the heat absorbed along i b f. Then it returns from f to i along a curved path, with sixty joules of work done on it. Find the heat exchanged. Pause the video and try all three parts. Done? Okay, let's check. Path i a f: heat is absorbed, so Q is plus two hundred; work is done by the system, so W is plus eighty. Delta U equals two hundred minus eighty: one hundred and twenty joules. Okay? So U f equals U i plus delta U: fifty plus one hundred and twenty, one hundred and seventy joules. Right? Now path i b f. Only the work is given here, thirty joules. So first tell me: along this path, what is delta U? The same, one hundred and twenty joules! Same start, same end. So Q equals delta U plus W: one hundred and twenty plus thirty, one hundred and fifty joules absorbed. See? Q changed with the path, but delta U did not. Clear? Last part: back from f to i, with sixty joules of work done on the system. What are the signs of delta U, and of W? Going back, the change is reversed; the arrow means from f to i. So delta U is minus one hundred and twenty joules. And work is done on the system, so W is minus sixty joules. Got it? So Q equals delta U plus W: minus one hundred and twenty, plus minus sixty, minus one hundred and eighty joules. The minus sign means the heat is released: one hundred and eighty joules leave the system. Two sign traps in one question, and you have handled both!

7:24Quick revision: Q = delta U + W, signs, cycles

So, quick revision. One: Q equals delta U plus W, okay? Heat in is stored, or spent as work done by the system. Two: heat absorbed is positive, heat released negative; work is positive in expansion, negative in compression. Three: delta U depends only on the end states, so in a cycle it is zero, and the net heat equals the net work. Now go and try the practice problems on this page. Keep your signs straight, and you will find them easy. All the best!