# Question e9c20

Dec 19, 2017

$\text{2.0 atm}$

#### Explanation:

Your tool of choice here will be the ideal gas law equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{P V = n R T}}}$

Here

• $P$ is the pressure of the gas
• $V$ is the volume it occupies
• $n$ is the number of moles of gas present in the sample
• $R$ is the universal gas constant, equal to $0.0821 \left(\text{atm L")/("mol K}\right)$
• $T$ is the absolute temperature of the gas

The first thing that you need to do here is to sue the ideal gas law equation to figure out how many moles of nitrogen gas are present in the initial sample.

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

${n}_{1} = \left(1.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 1.00 color(red)(cancel(color(black)("L"))))/(0.0821(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K}}}}\right)$

${n}_{1} = {\text{0.04085 moles N}}_{2}$

Now, you know that the number of moles of nitrogen gas present in the sample is increased by the addition of the $\text{2.00-g}$ sample, which contains

2.00 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28.0134color(red)(cancel(color(black)("g")))) = "0.07139 moles N"_2

Moreover, the temperature of the gas decreases from

${25}^{\circ} \text{C" + 273.15 = "298.15 K}$

to

$- {55}^{\circ} \text{C" + 273.15 = "218.15 K}$

So, if you write the ideal gas law equation for the initial state of the gas as

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

you can write the ideal gas law equation for the final state of the gas as

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

Divide these two equations to get

$\frac{{P}_{1} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}}{{P}_{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{V}}}} = \frac{{n}_{1} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot {T}_{1}}{{n}_{2} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{R}}} \cdot {T}_{2}}$

P_1/P_2 = (n_1 * T_1)/(n_2 * T_2

Rearrange to solve for ${P}_{2}$

${P}_{2} = {n}_{2} / {n}_{1} \cdot {T}_{2} / {T}_{1} \cdot {P}_{1}$

Here you have

${n}_{2} = \text{0.04085 moles" + "0.07139 moles" = "0.11224 moles}$

Plug in your values to find

P_2 = (0.11224 color(red)(cancel(color(black)("moles"))))/(0.04085color(red)(cancel(color(black)("moles")))) * (218.15 color(red)(cancel(color(black)("K"))))/(298.15color(red)(cancel(color(black)("K")))) * "1.00 atm"#

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{P}_{2} = \text{2.0 atm}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the two temperatures.