# Question 8dabc

Dec 16, 2014

The answer is : V = 2.80x10^-4L.

Gas' properties:
Volume (V)
Pressure (P)
Temperature (T- in Kelvins)
Amount (n - moles)
.
And a Constant R (0.0821 L x atm / mol x K)

The easiest way to solve this problem is to make a set up as follow: !!! STP !!!
V= ? L
P= 1 atm
T= 273 K
n= ?

1) We can get the amount of moles from $5.5 \cdot {10}^{- 4} g$ of $C {O}_{2}$ gas.

$M o l a r m a s {s}_{C {O}_{2}} = 12.01 + 2 \cdot 16 = 44.01 g$ of $C {O}_{2}$

Now take :

$5.5 \cdot {10}^{- 4} g C {O}_{2} \cdot \frac{1 m o l e C {O}_{2}}{44.01 g C {O}_{2}} = 1.25 \cdot {10}^{- 5}$ mol of $C {O}_{2}$

2) $P V = n R T$

$1 \cdot V = 1.25 \cdot {10}^{- 5} \cdot 0.0821 \cdot 273$
$V = 2.80 \cdot {10}^{- 4} L$

And that's it !

I strongly hope I was helpful !

David Tran
trananhdavid@gmail.com

Dec 16, 2014

You would use the ideal gas law in order to solve this problem. The equation for the ideal gas law is $\text{PV}$ = $\text{nRT}$. STP for the gas laws is $\text{0"^"o""C}$ and $\text{1 atm}$. The temperature must be converted to Kelvins, and the mass of ${\text{CO}}_{2}$ must be converted to moles.

Given/Known:
$\text{P}$ = $\text{1 atm}$
${\text{molar mass of CO}}_{2}$ = $\text{44.01 g/mol}$
$\text{n}$ = $\text{5.5 x 10"^(-4) "g}$ x $\text{1 mol CO2"/"44.01 g CO2}$ = ${\text{1.2 x 10"^(-5) "mol CO}}_{2}$
$\text{R}$ = $\text{0.08205746 L atm K"^(-1) "mol"^(-1)}$
$\text{T}$ = $\text{0"^"o""C" + 273.15 = 273.15 "K}$

Unknown:
$\text{V}$

Equation:
$\text{PV}$ = $\text{nRT}$

Solution: Divide both sides of the equation by $\text{P}$, to isolate $\text{V}$. Solve for $\text{V}$.

$\text{V}$ = $\text{nRT"/"P}$ = ${\text{1.2 x 10}}^{- 5}$x $\text{0.08205746}$ x $\text{273.15}$$/$$\text{1}$ = $\text{2.7 x 10"^(-4) "L}$ (Units removed in order to make the equation more compact.)

The volume of ${\text{5.5 x 10"^(-4) "g CO}}_{2}$ at STP is $\text{2.7 x 10"^(-4) "L}$.

Dec 16, 2014

An alternative approach to this problem is by using molar volume at STP.
We know that, at STP, one mole of any ideal gas occupies $22.4 L$.

The number of moles of $C {O}_{2}$, knowing that its molar mass is $44.01 \frac{g}{m o l}$, is

${n}_{C {O}_{2}} = {m}_{C {O}_{2}} / \left(m o l a r m a s s\right) = \frac{5.5 \cdot {10}^{- 4} g}{44.01 \frac{g}{m o l}} = 1.2 \cdot {10}^{- 5}$ moles

Therefore, since n = V/(V_(molar)#, we get

$V = {n}_{C {O}_{2}} \cdot {V}_{m o l a r} = 1.2 \cdot {10}^{- 5} m o l e s \cdot \frac{22.4 L}{1 m o l e} = 2.7 \cdot {10}^{- 4} L$

One can use this method as a primary tool or as a way to double-check the result determined using the ideal gas law.