# Question e5f52

Feb 18, 2015

Here's what I got.

#### Explanation:

In order to solve this problem, you must use the definition of partial pressure to determine the moles of each gas present in the mixture.

For a mixture of ideal gases, the partial pressure of each component is defined as the theoretical pressure that gas would exert if alone in the container.

So, in your case, the partial pressure of each of the three gases is equal to the total pressure that gas alone would exert in the same-volume container.

As a result, you can use this to determine how many moles of each gas you have by applying the ideal gas equation

$P V = n R T$.

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas
• $R$ is the universal gas constant, usually useful as $0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$
• $T$ is the absolute temperature of the gas

So,

${n}_{{N}_{2}} = \frac{{P}_{{N}_{2}} \cdot V}{R \cdot T} = \left(0.32 \cancel{\text{atm") * 2.5cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 15)cancel("K}}\right)$

$= {\text{0.0339 moles N}}_{2}$

n_("He") = (P_(He) * V)/(R * T) = (0.15cancel("atm") * 2.5cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 15)cancel("K"))

$= \text{ 0.0159 moles of He}$

n_("Ne") = (P_(Ne) * V)/(R * T) = (0.42cancel("atm") * 2.5cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 15)cancel("K"))

$= \text{ 0.0444 moles Ne}$

Partial pressure can also be expressed using a component's mole fraction, i.e. the ratio between the number of moles of the component and the total number of moles in the mixture.

${P}_{\text{partial") = chi_i * P_("total}}$

where

${\chi}_{i} = {n}_{i} / {n}_{\text{total}}$

The trick here is to recognize that the number of moles of each gas does not change when going from the initial container to the new one at STP. The new container will hold $0.0160$ moles of $\text{He}$ and $0.0444$ moles of $\text{Ne}$.

STP conditions are most commonly taken as a pressure is $\text{1 atm}$ and a temperature of $\text{273.15 K}$.

This means that the new partial pressures for $\text{Ne}$ and $\text{He}$ will be

P_("partial for Ne") = (0.0444 cancel("moles"))/((0.0444 + 0.0159)cancel("moles")) * "1 atm"

P_("partial for Ne") = "0.736 atm"

and

P_("partial for He") = (0.0159cancel("moles"))/((0.0444 + 0.0159)cancel("moles")) * "1 atm"

P_("partial for He") = "0.264 atm"

Finally, use these values, together with those for pressure and temperature stipulated by STP, to determine the new volume.

${V}_{\text{mixture") = ((n_("He") + n_("Ne")) * R * T)/P_("total}}$

V_("mixture") = ((0.0444 +0.0159)cancel("moles") * 0.082 (cancel("atm") * "L")/(cancel("mol") * cancel("K")) * 273.15cancel("K"))/(1cancel("atm"))

${V}_{\text{misture" = "1.4 L}}$

The answer is rounded to two sig figs.

Jan 17, 2017

By definition, the partial pressure of a component gas in a mixture of gases is the pressure exerted by that component when that gaseous component (in unaltered amount) occupies individually the toral volume of the gas mixture at the same temperature of the gas mixture.

Following this definition we can say that pressures of different components will be equal to their respective partial pressures when the components are separately kept $2.5 L$ closed vessel at ${15}^{\circ} C$

So removing ${N}_{2}$ we have

(1) initially

$H e \to 2.5 L \text{ under "0.15 atm" } {15}^{\circ} C = 288 K$

$N e \to 2.5 L \text{ under "0.42 atm" } {15}^{\circ} C = 288 K$

and (2) finally under $S T P$

$H e \to {V}_{\text{He"L " under "1 atm" }} {0}^{\circ} C = 273 K$

$N e \to {V}_{\text{Ne"L " under "1 atm" }} {15}^{\circ} C = 273 K$

Now by combined Boyle's and Charle's law we have the following relation

$\frac{{P}_{f} \times {V}_{f}}{T} _ f = \frac{{P}_{i} \times {V}_{i}}{T} _ i$

$\implies {V}_{f} = \frac{{P}_{i} \times {V}_{i} \times {T}_{f}}{{P}_{f} {T}_{i}}$
Here $i \to \text{initial" andf->"final}$

Applying this formula for remaining two gases we get final volumes at $S T P$ as follows

${V}_{\text{He}} = \frac{0.15 \times 2.5 \times 273}{1 \times 288}$

${V}_{\text{Ne}} = \frac{0.42 \times 2.5 \times 273}{1 \times 288}$

As both volumes are at $S T P$ they can be simply added to have the total volume of the gas mixture
Hence

${V}_{\text{Total}} = \frac{\left(0.15 + 0.42\right) \times 2.5 \times 273}{1 \times 288} = 1.4 L$ (2 sigfig)

Alternative

After removal of Nitrogen

The mixture has

$\text{initial volume } {V}_{i} = 2.5 L$

$\text{initial pressure } {P}_{i} = \left(0.15 + 0.42\right) = 0.57 a t m$

$\text{initial temperature } {T}_{i} = {15}^{\circ} C = 288 K$

"final volume "V_f=?#

$\text{final pressure } {P}_{f} = 1 a t m$

$\text{final temperature } {T}_{f} = {0}^{\circ} C = 273 K$

So ${V}_{f} = \frac{{P}_{i} \times {V}_{i} \times {T}_{f}}{{P}_{f} {T}_{i}}$

$\implies {V}_{f} = \frac{0.57 \times 2.5 \times 273}{1 \times 288} = 1.4 L$
(2sigfig)