# What volume will 5.2 xx 10^24 molecules of CO_2 occupy?

##### 2 Answers
Jan 14, 2018

Well, clearly a pellet of dry ice will occupy LESS volume than carbon dioxide gas....

#### Explanation:

And so we specify conditions of $P = 1 \cdot a t m$, and $T = 298 \cdot K$...and to answer your question MEANINGFULLY, we specify that $5.2 \times {10}^{24} \cdot \text{molecules of dry ice}$, were confined in a piston, and the gas expanded against $1 \cdot a t m \ldots .$

And so............ $V = \frac{n R T}{P} = \frac{\frac{5.2 \times {10}^{24} \cdot C {O}_{2}}{6.022 \times {10}^{23} \cdot m o {l}^{-} 1} \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 298 \cdot K}{1 \cdot a t m}$..

And I make this over $200 \cdot L$....

Jan 14, 2018

At what pressure and temperature?
Do you mean at STP?

#### Explanation:

We can use the ideal gas equation: PV = nRT
(Pressure)(Volume) = (moles)(gas constant)(Temperature)

You have given me these parameters:
moles = $\left(5.2 \cdot {10}^{24}\right) \cdot \frac{1 m o l}{6.022 \cdot {10}^{23}} = 8.635 m o l C {O}_{2}$
gas constant = $\frac{0.0821 \left(L\right) \left(a t m\right)}{\left(m o l\right) \left(K\right)}$
volume = x(variable)
pressure = ?(not given)
temperature = ?(not given)

If you mean at STP, where the pressure is 1 atm and the temperature is 273K, here is the equation and answer:
$\left(1 a t m\right) \left(x L\right) = \left(8.635 m o l C {O}_{2}\right) \left(\frac{0.0821 \cdot L \cdot a t m}{m o l \cdot K}\right) \left(273 K\right)$
$x \approx 193.5388$
Significant Digits:
color(blue)(x ~~ 190L