Question

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?

Answer 1

`1.24` `"atm"`

Explanation:

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

Using `n = frac(m)(M)`:

`Rightarrow n("O") = (frac(32.00)(2 times 15.99))` `"mol"`

`Rightarrow n("O") = (frac(32.00)(31.98))` `"mol"`

`therefore n("O") = 1.00` `"mol"`

Then, let's convert the units of the given temperature to `"K"`:

`Rightarrow T_((""^(@)"" "C")) = T_(("K")) - 273.15`

`Rightarrow 30.00 = T_(("K")) - 273.15`

`Rightarrow T_(("K")) = 303.15`

`therefore 30.00` `""^(@)"" "C"` `= 303.15` `"K"`

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

`Rightarrow P times 20.0` `"L"` `= 1.00` `"mol"` `times 0.0821` `"L atm mol"^(- 1) "K"^(- 1)` `times 303.15` `"K"`

`Rightarrow P times 20.0 = 1.00 times 0.0821` `"atm"` `times 303.15`

`Rightarrow P = (frac(0.0821 times 303.15)(20.0))` `"atm"`

`therefore P = 1.24` `"atm"`

Therefore, the pressure exerted by the oxygen gas is `1.24` `"atm"`.

Answer 2

`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:

`PV = nRT`

`P` is pressure (atm)
`V` is volume( liters)
`n` is moles of substance (`(Liters*atm)/(mols*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 = (nRT)/V`
Now, we find the rest of the numbers to plug in

`n`: We convert Grams of `O_"2"` to moles using dimensional analysis: `32.00g O_2 * (1 mol O_2)/(32.00g O_2) = 1 mol O_2`

`R`: We use the constant `0.0821`

`T`: We convert the temperature from Celcius to Fahrenheit, using the formula `K = ^oC + 273`
`K = 30 + 273 = 303K`

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

Now, we solve.

First by plugging in numbers:

`P = (nRT)/V`
=> `P = ((1 mol)(0.0821(Liters*atm)/(mols*K))(303K))/(20L)`

Which is the same as:

`P = ((1 mol)(0.0821(Liters*atm))(303K))/(20L*mols*K)`

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

`P = ((1)(0.0821 atm)(303))/(20)`

Now we simplify to get our answer:

`P = ((1)(0.0821 atm)(303))/(20)`

=> `P = (24.88atm)/(20)`

=> `P = 1.24 atm`

Answer 3

The pressure is 1.2 atm.

Explanation:

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

`PV=nRT`,

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 `"O"_2`, but the equation requires moles. Determine the mol `"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^@"C"` to Kelvins by adding `273.15`.

`30.00^@"C" + 273.15="303.15 K"`

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

Given/Known

`V="20. L"=2.0xx10^2color(white)(.)"L"`

`n="1.000 mol"`

`R="0.0821 L atm K"^(-1) "mol"^(-1)"`

`T="303.15 K"`

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

`P=(nRT)/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)