Question

What is the volume at STP of 3.44 x 1023 molecules of CO2

Answer 1

Answer:

C. 12.8 liters.

Explanation:

The Standard Temperature and Pressure (STP) of a gas are 273.15 K and 100 kilopascals. From Avogadro's Law, a mole of carbon dioxide contains $6.022 \times 10^{23}$ molecules. If we suppose that carbon dioxide behaves ideally, then the equation of state for ideal gas is:

$P\cdot V = n\cdot R_{u}\cdot T$ (1)

$P\cdot V = \frac{r\cdot R_{u}\cdot T}{N_{A}}$ (1b)

Where:

$P$ - Pressure, measured in pascals.

$V$ - Volume, measured in liters.

$r$ - Amount of molecules, no unit.

$N_{A}$ - Avogadro's number, no unit.

$R_{u}$ - Ideal gas constant, measured in pascal-liters per mole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 100000\,Pa$, $r = 3.44\times 10^{23}$, $N_{A} = 6.022\times 10^{23}$, $T = 273.15\,K$ and $R_{u} = 8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K}$, then the volume of carbon dioxide at STP is:

$V = \frac{r\cdot R_{u}\cdot T}{N_{A}\cdot P}$

$V = \frac{(3.44\times 10^{23})\cdot \left(8.314\times 10^{3}\,\frac{L\cdot Pa}{mol\cdot K} \right)\cdot (273.15\,K)}{(6.022\times 10^{23})\cdot (100000\,Pa)}$

$V = 12.972\,L$

Therefore, the correct answer is C.