Two cannisters of different sizes
contain the same gas at the same pressure and the same temperature. Cannister one contains 5.2 moles of
the gas, and cannister two contains 15.6 moles of the gas. What is the ratio of the volume of
cannister two to the volume of cannister one?
We’re told about these two
cannisters, cannister one and cannister two. We don’t yet know which cannister
is bigger than the other. But if we call the volume of
cannister two 𝑉 two and the volume of cannister one 𝑉 one, we want to solve for
this ratio: 𝑉 two divided by 𝑉 one. We’re told that both of these
cannisters contain the same gas. And we’re going to assume that this
gas is ideal. Under this assumption, we can use
the ideal gas law to describe this gas.
This law holds that the pressure of
an ideal gas times its volume is equal to the number of moles of the gas multiplied
by the molar gas constant times the gas temperature. Because we have two separate
cannisters, we can make two separate applications of the ideal gas law. For cannister one, we can write
that the gas’s pressure 𝑃 one times the container’s volume 𝑉 one equals the number
of moles of the gas in this container 𝑛 one times 𝑅 times the temperature of the
gas 𝑇 one.
The reason we haven’t written a one
subscript for the gas constant 𝑅 is because this is indeed a constant. Its value will remain the same
regardless of the properties of the gas. Writing a similar expression for
the gas in cannister two, we know that the pressure of this gas 𝑃 two times the
cannister’s volume 𝑉 two is equal to the number of moles of gas in cannister two 𝑛
two times the gas constant 𝑅 multiplied by the gas’s temperature in this cannister,
We’ve seen that we want to solve
for the ratio 𝑉 two divided by 𝑉 one. We can isolate the values of 𝑉 one
and 𝑉 two in these two respective equations. If we divide both sides of our
first equation by the pressure 𝑃 one, then that pressure cancels on the left. And we find that 𝑉 one equals 𝑛
one times 𝑅 times 𝑇 one all divided by 𝑃 one. Likewise, dividing our second
equation by 𝑃 two so that that factor cancels on the left, we find that 𝑉 two
equals 𝑛 two times 𝑅 times 𝑇 two divided by 𝑃 two.
Since we now have expressions both
for 𝑉 one and for 𝑉 two, we can substitute them into our ratio 𝑉 two divided by
𝑉 one. Clearing some space to work with
this fraction, we can multiply both numerator and denominator of this fraction by
one divided by the molar gas constant 𝑅. Doing so causes that constant to
cancel out of this expression entirely.
Next, we can recall some important
information given to us. We’re told that our two cannisters
contain gas which is at the same pressure and at the same temperature. In other words, what we’ve called
𝑃 one and 𝑃 two are equal, and what we’ve called 𝑇 one and 𝑇 two are also
equal. If we recognize this fact by
replacing 𝑃 one and 𝑃 two with 𝑃 and 𝑇 one and 𝑇 two with 𝑇, then we can see
that in both numerator and denominator of our equation, we’re multiplying by 𝑇
divided by 𝑃. If we choose then to multiply both
numerator and denominator by the inverse 𝑃 divided by 𝑇, then all of these values
will effectively cancel out. We’re left simply with the ratio of
moles of gas, 𝑛 two to 𝑛 one.
We are told that cannister one
contains 5.2 moles of gas, so 𝑛 one equals 5.2. Likewise, we know that cannister
two contains 15.6 moles of gas, so 𝑛 two is 15.6. Substituting these values in for 𝑛
two and 𝑛 one in our equation, we find that their ratio is equal to exactly
three. This then is the ratio of the
volume of cannister two to the volume of cannister one. Cannister two has three times as
much volume as cannister one.