# If "3.50 mols" of freon-12 gas was added to "7.48 mols" of it, and it started at "28.7 L", what is its final volume?

Jul 20, 2017

When you pay peanuts you get monkeys..........

${P}_{\text{initial}} \cong 20 \cdot L$

#### Explanation:

Generally, the answers are checked by experienced posters, and you will see many corrections and additions to answers. But this is not a cheat site; the emphasis is on the explanation not the final result, which the student is obligated to provide themselves. Learning will not occur otherwise.

Old Avogadro's Law states that $\text{equal volumes of all gases,}$ $\text{at the same temperature and pressure, have the same number}$ $\text{of molecules} .$

And thus $V \propto n$, where $n = \text{numbers of moles of gas}$

And so $\left(7.48 \cdot m o l + 3.50 \cdot m o l\right) \times k = 28.7 \cdot L$

$k = 2.61 \cdot L \cdot m o {l}^{-} 1$.

And so ${P}_{\text{initial}} = 2.61 \cdot L \cdot m o {l}^{-} 1 \times 7.48 \cdot m o l \cong 20 \cdot L$.

Jul 20, 2017

${V}_{1} = \text{19.6 L}$

That is, if freon-12, or ${\text{CF"_2"Cl}}_{2}$, at whatever the balloon temperature and pressure are, is ideal...

Well, if we assume the Freon-12 gas is ideal, then we can use the ideal gas law:

$P V = n R T$,

where:

• $P$ is the pressure in, say, $\text{atm}$.
• $V$ is the volume, usually in $\text{L}$.
• $n$ is the mols of ideal gas.
• $R = \text{0.082057 L"cdot"atm/mol"cdot"K}$ is the universal gas constant, if pressure is in $\text{atm}$ and volume is in $\text{L}$.
• $T$ is temperature in $\text{K}$.

There were unstated conditions, and they were:

• constant temperature
• constant pressure

Only then can we come up with Avogadro's principle... We set initial and final states as:

$P {V}_{1} = {n}_{1} R T$

$P {V}_{2} = {n}_{2} R T$

And if we rearrange, we realize that...

${V}_{1} / {n}_{1} = {V}_{2} / {n}_{2} = \frac{R T}{P}$

i.e. that the molar volume of the freon gas remains constant (as it must, if it is to remain the same gas at the same temperature and pressure).

So, if we add $\text{3.50 mols}$ of freon gas to $\text{7.48 mols}$, we obtain:

${n}_{2} = {\overbrace{\text{7.48 mols")^(n_1) + overbrace("3.50 mols}}}^{\Delta n}$

or that ${n}_{2} = \text{10.98 mols}$. Thus, the initial volume is:

$\textcolor{b l u e}{{V}_{1}} = {n}_{1} / {n}_{2} \times {V}_{2}$

$= \text{7.48 mols"/"10.98 mols" xx "28.7 L}$

$=$ $\textcolor{b l u e}{\text{19.6 L}}$ to three sig figs.

And as always, we should check that our answer makes physical sense.

We added more mols of gas at constant temperature and pressure, so the balloon SHOULD expand, directly proportional to the increase in mols of gas... and it does!

V_2/V_1 -= "28.7 L"/"19.6 L" stackrel(?" ")(=) ("10.98 mols")/("7.48 mols") -= n_2/n_1

$1.464 \approx 1.468$ color(blue)(sqrt"")

But of course, this calculation is only valid for ideal gases. We are not given the temperature or pressure, so if we are in an evacuated refrigerator, we could very well be staring at one of the worst examples of an ideal gas...