# Question ffdb3

Feb 20, 2017

#### Answer:

A) $\text{3.47 L CO"_2}$

B) $\text{1.23 L CO"_2}$

C) $\text{2.13 L CO"_2}$

#### Explanation:

Use the ideal gas law:

"PV=nRT

where $P$ is pressure, $V$ is volume, $n$ is amount in moles, $R$ is the gas constant, which varies depending on the pressure unit and other values, and $T$, which is the Kelvin temperature.

Since the temperature is required to be in Kelvins, the Celsius temperature must be converted to Kelvins by adding $273.15$.
You can find the values for the gas constant at http://www.cpp.edu/~lllee/gasconstant.pdf

${\textcolor{p u r p \le}{\text{MOLES CO}}}_{2}$

Before starting, the mass of carbon dioxide gas must be converted to moles. Divide the given mass by the molar mass of $\text{CO"_2}$.
https://www.ncbi.nlm.nih.gov/pccompound?term=CO2

(6.73cancel"g")/(44.009cancel"g"/"mol")=color(purple)"0.153 mol CO"_2"

$\textcolor{red}{\text{QUESTION A:}}$ Conditions: $\text{STP}$

Modern $\text{STP}$ temperature is ${0}^{\circ} \text{C}$ or $\text{273.15 K}$, and pressure is $\text{100 kPa}$.

$R = \left(8.3144621 \textcolor{w h i t e}{.} \text{L"xx "kPa")/("K"xx"mol}\right)$

Rearrange the equation to isolate volume, $V$.

$V = \frac{n R T}{P}$

$V = \left(0.153 \cancel{\text{mol"xx8.3144621color(white)(.)"L"xxcancel"kPa"xx273.15cancel"K")/(cancel"K"xx cancel"mol"xx100cancel"kPa")=color(red)("3.47 L CO"_2}}\right)$ rounded to three significant figures

$\textcolor{b l u e}{\text{QUESTION B:}}$ Conditions: ${120}^{\circ} \text{C}$ and $\text{4.00 atm}$

Convert Celsius temperature to Kelvins.

${120}^{\circ} \text{C" +273.15="393 K}$

$R = \left(0.08205746 \text{L"xx"atm")/("K"xx"mol}\right)$

Solve for $V$.

$V = \frac{n R T}{P}$

$V = \left(0.153 \cancel{\text{mol"xx0.08205746color(white)(.)"L"xxcancel"atm"xx393cancel"K")/(cancel"K"xxcancel"mol"xx4.00cancel"atm")=color(blue)("1.23 L CO"_2}}\right)$

$\textcolor{g r e e n}{\text{QUESTION C:}}$ Conditions: $\text{209 K}$ and $\text{125 kPa}$

$R = \left(8.3144621 \textcolor{w h i t e}{.} \text{L"xx "kPa")/("K"xx"mol}\right)$

Rearrange the equation to isolate volume, $V$.

$V = \frac{n R T}{P}$

V=(0.153cancel"mol"xx8.3144621color(white)(.) "L"xx 209cancel"K")/(cancel"K"xxcancel"mol"xx125cancel"kPa")=color(green)"2.13 L CO"_2"# rounded to three significant figures