# Which flask contains gas of molar mass 30, and which contains gas of molar mass 60?

## Suppose you are given two 1-L flasks and told that one contains a gas of molar mass 30, the other a gas of molar mass 60, both at the same temperature. The pressure in flask A is X kPa, and the mass of gas in the flask is 1.2 g. The pressure in flask B is 0.5 X kPa, and the mass of gas in that flask is 1.2 g.

Aug 11, 2016

Here's what I got.

#### Explanation:

!! QUICK ANSWER !!

The trick here is to realize that

• pressure is directly proportional to the number of moles of gas present in each flask
• the number of moles is inversely proportional to the molar mass of the gas

Pressure is directly proportional to the number of moles of gas, which means that a pressure that is twice as high corresponds to twice as many moles of gas present in the flask.

Therefore, flask A contains twice as many moles of gas as flask B. Now, the heavier gas will contain fewer moles in the same mass.

The two samples have the same mass, but flask A contains twice as many moles as flask B, which can only mean that the gas present in flask A has a molar mass of ${\text{30 g mol}}^{- 1}$ and the gas present in flask B has a molar mass of ${\text{60 g mol}}^{- 1}$.

Therefore,

${\text{Flask A " -> " 30 g mol}}^{- 1}$

${\text{Flask B " -> " 60 g mol}}^{- 1}$

$\textcolor{w h i t e}{\frac{a}{a}}$

!! DETAILED EXPLANATION !!

The problem tells you that the two flasks have the same volume, let's say $V$, and that they are kept at the same temperature, let's say $T$.

This means that if you start from the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant
$T$ - the absolute temperature of the gas

you can say that you have

${P}_{A} \cdot V = {n}_{A} \cdot R \cdot T \to$ for the gas present in flask A

${P}_{b} \cdot V = {n}_{B} \cdot R \cdot T \to$ for the gas present in flask B

Divide these two equations to find a relationship between the pressure of the two gases and the number of moles of gas present in each flask

$\frac{{P}_{A} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}{{P}_{B} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}} = \frac{{N}_{A} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{T}}}}{{n}_{B} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{T}}}} \implies {P}_{A} / {P}_{B} = {n}_{A} / {n}_{B}$

Now, you know that the pressure in flask A, given as $\text{X kPa}$, is twice as high as the pressure in flask B, given as (1/2 * "X") color(white)(a)"kPa".

This means that the ratio between the number of moles of gas present in each flask will be

$\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{\text{X")))color(red)(cancel(color(black)("kPa"))))/(1/2 * color(red)(cancel(color(black)("X"))) color(red)(cancel(color(black)("kPa}}}}\right) = {n}_{A} / {n}_{B} \implies {n}_{A} / {n}_{B} = 2$

This is equivalent to

color(purple)(|bar(ul(color(white)(a/a)color(black)(n_A = 2 * n_B)color(white)(a/a)|)))" " " "color(orange)("(*)")

You now know that flask A contains twice as many moles of gas as flask B.

Now, you know that the two gases have the same mass, given as $\text{1.2 g}$. In order for the sample to have the same mass, you need the molar mass of the gas present in flask B, let's say ${M}_{B}$, to be twice as big as the molar mass of the gas present in flask A, let's say ${M}_{A}$.

This is the case because

n_A = (1.2 color(red)(cancel(color(black)("g"))) )/(M_A color(red)(cancel(color(black)("g"))) "mol"^(-1)) = 1.2/M_Acolor(white)(a)"moles"

n_B = (1.2 color(red)(cancel(color(black)("g"))))/(M_B color(red)(cancel(color(black)("g"))) "mol"^(-1)) = 1.2/M_B color(white)(a)"moles"

According to equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$, you will have

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{1.2}}}}{M} _ A = 2 \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{1.2}}}}{M} _ B \implies \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{M}_{B} = 2 \cdot {M}_{A}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that flask A contains the gas with the molar mass of ${\text{30 g mol}}^{- 1}$ and flask B contains the gas with the molar mass of ${\text{60 g mol}}^{- 1}$.

Once again,

${\text{Flask A " -> " 30 g mol}}^{- 1}$

${\text{Flask B " -> " 60 g mol}}^{- 1}$