# What is the pressure exerted by 32.00 g of Oxygen gas in a 20. L container at 30.00 C? Use R = 0.0821 L atm/mol k?

May 20, 2017

$1.24$ $\text{atm}$

#### Explanation:

First, let's determine the number of mole of oxygen gas.

Using $n = \frac{m}{M}$:

$R i g h t a r r o w n \left(\text{O}\right) = \left(\frac{32.00}{2 \times 15.99}\right)$ $\text{mol}$

$R i g h t a r r o w n \left(\text{O}\right) = \left(\frac{32.00}{31.98}\right)$ $\text{mol}$

$\therefore n \left(\text{O}\right) = 1.00$ $\text{mol}$

Then, let's convert the units of the given temperature to $\text{K}$:

$R i g h t a r r o w {T}_{\left(\text{^(@)"" "C")) = T_(("K}\right)} - 273.15$

$R i g h t a r r o w 30.00 = {T}_{\left(\text{K}\right)} - 273.15$

$R i g h t a r r o w {T}_{\left(\text{K}\right)} = 303.15$

$\therefore 30.00$ $\text{^(@)"" "C}$ $= 303.15$ $\text{K}$

Now, let's substitute all relevant values into the equation $P V = n R T$:

$R i g h t a r r o w P \times 20.0$ $\text{L}$ $= 1.00$ $\text{mol}$ $\times 0.0821$ ${\text{L atm mol"^(- 1) "K}}^{- 1}$ $\times 303.15$ $\text{K}$

$R i g h t a r r o w P \times 20.0 = 1.00 \times 0.0821$ $\text{atm}$ $\times 303.15$

$R i g h t a r r o w P = \left(\frac{0.0821 \times 303.15}{20.0}\right)$ $\text{atm}$

$\therefore P = 1.24$ $\text{atm}$

Therefore, the pressure exerted by the oxygen gas is $1.24$ $\text{atm}$.

May 20, 2017

$1.24$ atm, rounded to two significant figures

#### Explanation:

Note: mols and moles are the same things; formatting issues prevent me from using the same word throughout the answer
We will use the ideal gas equation, which is as follows:

$P V = n R T$

$P$ is pressure (atm)
$V$ is volume( liters)
$n$ is moles of substance ($\frac{L i t e r s \cdot a t m}{m o l s \cdot K}$)
$R$ is the constant $0.0821$
$T$ is temperature (kelvin).

We are solving for $P$ in this equation, so the formula is reworked as so:
$P = \frac{n R T}{V}$
Now, we find the rest of the numbers to plug in

$n$: We convert Grams of ${O}_{\text{2}}$ to moles using dimensional analysis: $32.00 g {O}_{2} \cdot \frac{1 m o l {O}_{2}}{32.00 g {O}_{2}} = 1 m o l {O}_{2}$

$R$: We use the constant $0.0821$

$T$: We convert the temperature from Celcius to Fahrenheit, using the formula $K {=}^{o} C + 273$
$K = 30 + 273 = 303 K$

$V$: We use the given liters, which is $20 L$

Now, we solve.

First by plugging in numbers:

$P = \frac{n R T}{V}$
=> $P = \frac{\left(1 m o l\right) \left(0.0821 \frac{L i t e r s \cdot a t m}{m o l s \cdot K}\right) \left(303 K\right)}{20 L}$

Which is the same as:

$P = \frac{\left(1 m o l\right) \left(0.0821 \left(L i t e r s \cdot a t m\right)\right) \left(303 K\right)}{20 L \cdot m o l s \cdot K}$

Liters, moles, and Kelvin cancel out, giving us:

$P = \frac{\left(1\right) \left(0.0821 a t m\right) \left(303\right)}{20}$

Now we simplify to get our answer:

$P = \frac{\left(1\right) \left(0.0821 a t m\right) \left(303\right)}{20}$

=> $P = \frac{24.88 a t m}{20}$

=> $P = 1.24 a t m$

May 20, 2017

The pressure is 1.2 atm.

#### Explanation:

This is an ideal gas law problem. The equation is:

$P V = n R T$,

where $P$ is pressure, $V$ is volume, $n$ is moles, $R$ is a gas constant, and $T$ is temperature in Kelvins.

You have been given the mass of ${\text{O}}_{2}$, but the equation requires moles. Determine the mol ${\text{O}}_{2}$ by multiplying its given mass by the inverse of its molar mass (31.998 g/mol).

32.00color(red)cancel(color(black)("g O"_2))xx(1"mol O"_2)/(31.998color(red)cancel(color(black)("g O"_2)))="1.000 mol O"_2"

You have been given temperature in degrees Celsius, but gas problems require the temperature to be in Kelvins. Convert ${30.00}^{\circ} \text{C}$ to Kelvins by adding $273.15$.

${30.00}^{\circ} \text{C" + 273.15="303.15 K}$

color(blue)("Now organize your data."

Given/Known

$V = \text{20. L"=2.0xx10^2color(white)(.)"L}$

$n = \text{1.000 mol}$

$R = \text{0.0821 L atm K"^(-1) "mol"^(-1)}$

$T = \text{303.15 K}$

color(blue)("Solution."
Rearrange the equation to isolate $P$. Insert your data and solve.

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

P=(1.000color(red)cancel(color(black)("mol"))xx0.0821color(red)cancel(color(black)("L")) "atm" color(red)cancel(color(black)("K"))^(-1) color(red)cancel(color(black)("mol"))^(-1)xx303.15color(red)cancel(color(black)("K")))/(2.0xx10^2color(red)cancel(color(black)("L")))="1.2 atm" rounded to two significant figures)