# Question 32207

Feb 17, 2017

$\text{680 torr}$

#### Explanation:

The idea here is that the pressure of a gas is directly proportional to the amount of gas present when volume and temperature are kept constant.

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P \propto n}}} \to$ when volume and temperature are kept constant

This can be expressed as

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{P}_{1} / {n}_{1} = {P}_{2} / {n}_{2}}}}$

Here

• ${P}_{1}$, ${n}_{1}$ are the pressure and number of moles of gas at an initial state
• ${P}_{2}$, ${n}_{2}$ are the pressure and number of moles of gas at a final state

Now, you don't have to convert the sample from grams to moles because the number of moles of a substance is proportional to its mass as given by

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{n = \frac{m}{M} _ M}}}$

Here

• $m$ is the mass of the gas
• ${M}_{M}$ is the molar mass of the gas

So, you know that you start with $\text{35 g}$ of ethylene. If you take ${M}_{M}$ to be the molar mass of ethylene, you can say that you start with

${n}_{1} = \frac{35}{M} _ M$

After you remove $\text{5 g}$ of ethylene, the equivalent of $\frac{5}{M} _ M$ moles, you will be left with

${n}_{2} = \frac{35}{M} _ M - \frac{5}{M} _ M = \frac{30}{M} _ M$

moles of ethylene in the container. This means that you will have

${P}_{1} / \left(\frac{35}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{M}_{M}}}}}\right) = {P}_{2} / \left(\frac{30}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{M}_{M}}}}}\right)$

which is equivalent to

${P}_{2} = \frac{30}{35} \cdot {P}_{1}$

Plug in your value to find

P_2 = 30/35 * "793 torr" = color(darkgreen)(ul(color(black)("680 torr")))#

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of gas removed from the container.

So, does the result make sense?

Since pressure is proportional to the amount of gas present in the container, decreasing the amount of gas will cause the pressure to decrease as well.