Question #ffdb3

1 Answer
Feb 20, 2017

A) "3.47 L CO"_2"

B) "1.23 L CO"_2"

C) "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

color(purple)"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 "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"

color(red)"QUESTION A:" Conditions: "STP"

Modern "STP" temperature is 0^@"C" or "273.15 K", and pressure is "100 kPa".

R=(8.3144621color(white)(.) "L"xx "kPa")/("K"xx"mol")

Rearrange the equation to isolate volume, V.

V=(nRT)/P

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

color(blue)("QUESTION B:") Conditions: 120^@"C" and "4.00 atm"

Convert Celsius temperature to Kelvins.

120^@"C" +273.15="393 K"

R=(0.08205746"L"xx"atm")/("K"xx"mol")

Solve for V.

V=(nRT)/P

V=(0.153cancel"mol"xx0.08205746color(white)(.)"L"xxcancel"atm"xx393cancel"K")/(cancel"K"xxcancel"mol"xx4.00cancel"atm")=color(blue)("1.23 L CO"_2")

color(green)"QUESTION C:" Conditions: "209 K" and "125 kPa"

R=(8.3144621color(white)(.) "L"xx "kPa")/("K"xx"mol")

Rearrange the equation to isolate volume, V.

V=(nRT)/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