Constant-volume calorimetry | Thermodynamics | AP Chemistry | Khan Academy

Calorimetry refers to the measurement of heat flow, and there are many different types of calorimeters. In this case, we're looking at a constant volume calorimeter, which is also called a bomb calorimeter. Let's look at how a bomb calorimeter works.

First, the sample to be combusted is placed in a container that has some oxygen, and then there are some ignition wires that go into this little container here. The sample is ignited, and heat is given off by the combustion reaction. So, heat is being transferred from our sample to the water in our container.

Let me go ahead and draw some water in here. So, imagine we have some water, and it's important to note that the container is rigid. The walls are very solid and cannot move. There's also something to stir the water, and since heat is being transferred from the combustion reaction to the water, the temperature of the water will increase, which we can see on the thermometer.

Now that we understand how a bomb calorimeter works, let's think about that heat that's being transferred from the combustion reaction to the water. So, that heat is q. Let's go back to the first law of thermodynamics, which says that the change in the internal energy of the system is equal to q plus w, where q is the heat that's transferred and w is the work done.

Let's say we do this combustion reaction in a container with a movable piston, and the combustion reaction is performed under the constant pressure of the atmosphere. So, this time when we do the combustion reaction, we will transfer some heat. Heat is being transferred from the combustion reaction, and we would also produce some gases, which would push up on the pistons, and so the piston would move up.

Since the piston is moving, work is being done by the combustion reaction. In this case, the heat that's transferred, q, is done under constant pressure, and so we can write qp here. By definition, the heat that's transferred at constant pressure is the change in the enthalpy, Δh.

For this example with the container with the movable piston, when we did our combustion reaction, the heat that's transferred at constant pressure is equal to the enthalpy, Δh. As the gases expand and push in the piston, work is done.

Let's compare the example with the movable piston to our bomb calorimeter. Our bomb calorimeter has rigid walls, and therefore no work can be done. So, the work done is equal to zero. When we plug that into the first law of thermodynamics, we find that the change in the internal energy, ΔE, is equal to the heat transferred, q.

Since this is a constant volume calorimeter, right, the walls are rigid, we can write q sub v here. So, this heat that's transferred from our combustion reaction, in this case, is not equal to the change in the enthalpy; it's equal to the change in the internal energy, ΔE.

So, the heat that's transferred at constant pressure is equal to the change in the enthalpy, Δh, while the heat that's transferred at constant volume is equal to the change in the internal energy, ΔE.

To do a constant volume calorimetry problem, we need to know the heat capacity of the calorimeter, which is symbolized by c with a subscript cal. To find the heat capacity of the calorimeter, we need to combust something that we know the exact amount of heat for. For example, if you combust exactly one gram of benzoic acid, you'll get 26.38 kilojoules released of energy.

So let's say we have a 0.2350 gram sample of benzoic acid, and we put that in our calorimeter, and we go ahead and combust the benzoic acid. We find that the temperature increases by positive 1.642 degrees Celsius.

To find the heat capacity for the calorimeter, first, we take our known amount, which was 26.38 kilojoules per gram, and we multiply that by how much benzoic acid we used in our calorimeter, which was 0.2350 grams. So, the grams will cancel out, and this is equal to 6.199 kilojoules. Next, we divide this by our temperature change, which was positive 1.65 degrees Celsius, and this gives us the heat capacity of our calorimeter, which turns out to be 3.775 kilojoules per degree Celsius.

Now that we know the heat capacity for our specific calorimeter, we can use this value to calculate the heat of combustion for another substance. So, the heat of combustion for another substance, or just q, would be equal to the negative of the heat capacity of the calorimeter times the change in the temperature of the water in that calorimeter.

Let's say our goal is to calculate the heat of combustion of caffeine in kilojoules per mole. We take 0.265 grams of caffeine, we put that in our calorimeter, we combust it, and we find the temperature of the water increases by positive 1.525 degrees Celsius.

So to calculate q, q is equal to the negative of the heat capacity of the calorimeter, which is 3.775 kilojoules per degree Celsius, and we multiply that by the temperature change, which is 1.525 degrees Celsius. So, degrees Celsius cancels out, and this gives us negative 5.757 kilojoules.

Technically, this is the heat transfer at constant volume, so we can even write q sub v in here. Remember, this is equal to the change in the internal energy of our system, so qv is equal to ΔE. Since our goal is to find the heat of combustion of caffeine in kilojoules per mole, next we need to take our grams of caffeine, which is 0.265 grams, and divide that by the molar mass of caffeine.

So, grams will cancel out and give us moles of caffeine. This calculation is equal to 1.36 times 10 to the negative third moles of caffeine. Now all we have to do is divide our heat, which is negative 5.757 kilojoules, by our moles of 1.36 times 10 to the negative third moles to give us a final value of negative 4.23 times 10 to the third kilojoules per mole, with the negative sign meaning heat is given off.

So we can say that this is the value; this is the change in the internal energy for our reaction in kilojoules per mole. Often, the change in enthalpy is about the same as the change in the internal energy, so we can say that this is approximately equal to the change in the enthalpy for the reaction as well.