Question #ffdb3

1 Answer
Feb 20, 2017

Answer:

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