### Video Transcript

A gas with a mass of 0.125 kilograms is contained in a rigid container with a volume
of 2.5 cubic meters. The gas is heated, increasing its temperature by 35 Kelvin and also increasing its
pressure as shown in the diagram. Calculate the change in the internal energy of the gas due to its heating. Use a value of 1.75 kilojoules per kilogram Kelvin for the specific heat capacity of
a constant volume of the gas. Assume the heating of the container is negligible. Answer to three significant figures.

Our diagram, which is a pressure versus volume diagram of our gas, shows that the
volume of the gas is fixed at 2.5 cubic meters while the pressure increases from
2500 to 4500 pascals. Knowing this as well as some other information about the gas, we want to calculate
the change in the internal energy of the gas due to this heating. To do that, we’ll clear some space on our board and we’ll also record the information
given to us about this gas. For our variables, we’ve chosen to use 𝑚 for the gas’s mass, 𝑉 for its volume, Δ𝑡
for its change in temperature, and 𝐶 sub 𝑣 for its specific heat capacity at
constant volume.

We want to solve for the change in the gas’s internal energy. We can call that Δ𝑢. And to do that, we’re going to rely on the information we’ve been given in the
problem statement. Let’s start with 𝐶 sub 𝑣, the specific heat capacity of the gas at constant
volume. This tells us how much energy it takes to increase the temperature of one kilogram of
our gas. Here’s the good thing! Because we know the change in temperature of our gas as well as its mass, we can use
both those values to solve for the total energy put into the gas. The way we’ll find that is by multiplying our specific heat capacity at constant
volume by the gas’s mass and its change in temperature.

Notice that as we carry out this multiplication, the units of kilograms cancel from
this expression as do the units of Kelvin, temperature. When we multiply these three numbers together, we’re going to find the total energy
that’s put into the gas through heating. Now often, some of that energy would be expressed as an expansion of the gas in
volume. But in this case, that doesn’t happen because the gas is confined to a fixed
volume. That means that all of this energy, which is given to the gas through heating, goes
to raising its internal energy.

In other words, Δ𝑢, the variable we want to solve for, is equal to the product of
these three numbers. And again that’s because the energy in the gas has nowhere else to go. It can’t expand because it’s in a fixed container. As one last step before we find this product, let’s convert this energy 1.75
kilojoules into units of joules. Because 1000 joules is equal to one kilojoule, that means that 1.75 kilojoules can be
rewritten as 1.75 times 10 to the third joules. Finally, we multiply the numbers together and, to three significant figures, find a
result of 7660 joules. That is the change in the internal energy of this gas due to heating.