Video Transcript
A gas is heated from 19 degrees
Celsius to 80 degrees Celsius, while the volume is kept constant. If the initial pressure was 2000
pascals, what is the pressure of the gas after it has been heated? Give your answer to three
significant figures.
Alright, so in this example, we
have a gas which starts out at 19 degrees Celsius, but then it’s heated and its
temperature goes up to 80 degrees Celsius. We’re told when the temperature of
the gas is 19 degrees Celsius, it has a pressure of 2000 pascals. Then as the temperature of the gas
changes, we want to see what impact — if any — this has on the pressure of the
gas. As we consider what kind of method
would help us figure out this final gas pressure, we find a clue in the problem
statement: this gas was heated and the volume was kept constant.
When the volume of a gas is held
constant over some process, that means that Gay-Lussac’s law applies to that
process. This law tells us that for a gas
held at constant volume, the pressure of that gas divided by its temperature is
equal to a constant value. In other words, even as the
pressure or the temperature of the gas will change, so long as the volume of the gas
is constant, that ratio 𝑃 divided by 𝑇 is always the same.
Another way of writing this is to
consider some change that happens in a gas and to say that the pressure divided by
the temperature before the change 𝑃 one divided by 𝑇 one is equal to the pressure
divided by temperature after the change. This second way of writing the law
is especially helpful for us because we indeed have a gas that has gone through a
change. It’s been heated. So we can say the initial pressure
of our gas divided by its initial temperature is equal to its final pressure —
that’s what we want to solve for — divided by its final temperature.
Since it’s 𝑃 two, the final gas
pressure that we want to solve for, let’s rearrange this equation to let us do
that. If we multiply both sides of the
equation by 𝑇 two, the temperature of the gas after it’s heated, then that term
cancels on the right-hand side and we have 𝑃 two all by itself. We can write the resulting equation
this way. We can say that 𝑃 two, the final
pressure of the gas, equals 𝑃 one times the temperature ratio 𝑇 two divided by 𝑇
one.
Now at this point in our solution,
it’s important to notice something. See the temperatures we’ve been
given in the problem statement, 19 degrees Celsius; 80 degrees Celsius. That is these temperatures are
given as values on the Celsius scale. But that’s not the SI unit of
temperature. The SI temperature unit is
kelvin. In other words, before we plug in
𝑇 two and 𝑇 one for our calculation, we want to convert these given temperatures
in to temperatures on the Kelvin scale. If we forgot to do that, our answer
would be off.
So we want to know how to convert a
temperature from degrees Celsius to kelvin. And roughly speaking, a temperature
in kelvin is equal to that same temperature in degrees Celsius plus 273. So for example, if we had a
temperature in Celsius of zero degrees, that is freezing, then the equivalent
temperature in kelvin would be zero plus 273 or 273 kelvin. Using this relationship, let’s
write out the initial and final temperatures of our gas in kelvin. 𝑇 one, the initial temperature of
the gas, will be equal to 19 plus 273 and all this will be in kelvin. All that together is 292
kelvin. This value in kelvin is the value
we’ll use to substitute in for 𝑇 one in our equation.
Next, what about 𝑇 two, the
temperature of our gas after it’s been heated? That’s equal to 80 plus 273 all in
kelvin. And this equals 353 kelvin. That’s the temperature we’ll use
for 𝑇 two in our equation. With these values plugged in,
notice that the units kelvin cancel out. But be careful. This doesn’t mean that if we had
used degrees Celsius instead of kelvin, the units would also have cancelled and
therefore wouldn’t have made a difference. It’s true that if we had forgotten
to convert our initial temperatures in degrees Celsius into kelvin and simply plug
them into the equation, then those units would have also cancelled.
But notice what our fraction would
have been. Instead of 353 divided by 292, we
would have had a fraction of 80 divided by 19. These two fractions are not the
same thing. All that to say whenever we’re
working within the SI system as we usually are, then whenever an equation involves
temperature, we’ll want to make sure to use the Kelvin temperature scale in our
calculation. Okay, we’re almost ready to solve
for 𝑃 two, the final pressure of this gas.
In order to do so, let’s plug in
for 𝑃 one, the initial pressure. And that’s given as 2000
pascals. This unit, pascals, is the base
unit for pressure in the SI system. As the unit of pressure, a pascal
is equal to a newton of force divided by a square metre of area. So when we see 2000 pascals, that’s
another way of saying 2000 newtons of force per square metre of area. And that value can now go in for 𝑃
one in our equation. We’re now ready to calculate 𝑃
two, the pressure of the gas after it’s been heated. We see the units of this result
will be pascals. And when we calculate this
expression and round it to three significant figures, we find a result of 2420
pascals. This is the pressure of the gas
after it’s been heated.
