# Question 0f7fb

Oct 1, 2015

$\text{22.7 L}$

#### Explanation:

The idea here is that you first need to sue the ideal gas law to find how many moles of oxygen gas you have in that sample, then use the combined gas law to find the volume of the sample at STP conditions.

So, start by calculating the number of moles of oxygen gas you have in that sample - do not forget to convert the pressure from torr to atm, the volume from lililiters to liters, and the temperature to Kelvin

$P V = n R T \implies n = \frac{P V}{R T}$

n = (760/738.2color(red)(cancel(color(black)("atm"))) * 380 * 10^(-3)color(red)(cancel(color(black)("L"))))/(0.082(color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "0.0160 moles O"""_2

Now, STP conditions imply a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$. If you keep the number of moles of gas constant, you can say that

${P}_{1} \cdot {V}_{1} = n R \cdot {T}_{1} \to$ for your initial conditions

${P}_{2} \cdot {V}_{2} = n R \cdot {T}_{2} \to$ for STP conditions

Divide these two equations to get

$\frac{{P}_{1} \cdot {V}_{1}}{{P}_{2} \cdot {V}_{2}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R}}} \cdot {T}_{1}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{n \cdot R}}} \cdot {T}_{2}}$

which is equivalent to the combined gas law form

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

Rearrange this equation to solve for ${V}_{2}$, the volume the sample of oxygen would occupy at STP

${V}_{2} = {P}_{1} / {P}_{2} \cdot {T}_{2} / {T}_{1} \cdot {V}_{1}$

This means that you have

V_2 = (760/738.2color(red)(cancel(color(black)("atm"))))/(100/101.325color(red)(cancel(color(black)("atm")))) * (273.15color(red)(cancel(color(black)("K"))))/((273.15 + 25)color(red)(cancel(color(black)("K")))) * 380 * 10^(-3) "L"

${V}_{2} = \text{0.3631668 L}$

To get the molar volume of oxygen gas at STP, divide this value by the number of moles you have in the sample

${V}_{\text{molar STP" = V_"STP}} / n$

${V}_{\text{molar STP" = "0.3631668 L"/"0.0160 moles" = "22.697 L/mol}}$

I'' leave the answer rounded to three sig figs

V_"molar STP" = color(green)("22.7 L/mol")#

SIDE NOTE Many online sources and textbooks still define STP conditions as a pressure of 1 atm and a temperature of ${0}^{\circ} \text{C}$.

That is a very old definition of STP conditions that will produce a molar volume of 22.4 L. If you are supposed to use those conditions for pressure and temperature, simply redo the calculations using 1 atm instead of 100 kPa.