Question

At what temperature in Celsius will 19.4 g of molecular Ozone exert a pressure of 1820 mmHg in a 5.12 L cylinder?

Answer 1

`T` = `369` K

Explanation:

Ozone = `19.4`g
Molar mass = `48` g/mol
No. of moles = `19.4` g/`48`g.mol-1 = `0.4041` moles
Pressure = `1820` mmHg
P = `1820` mmHg/`760` mm Hg = 2.39 atm
Volume = 5.12 L
R = 0.08206 atm.L.K-1.mol-1
According to ideal gas law
`PV` = `nRT`
`T` = `PV`/`nR`

`T` = (`2.39 atm` x `5.12L`) / (`0.4041` moles x `0.08206 atm.L.K-1.mol-1`)

`T` = `369` K

Answer 2

The temperature in degrees Celsius is `97^@"C"`.

Explanation:

You will need to use the ideal gas law in order to answer this question. The formula is:

`PV=nRT`,

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

Determine Moles `"O"_3`

In order to determine the moles of molecular ozone `("O"_3")`, you need to divide its given mass by its molar mass (g/mol), which is `"47.9997 g/mol O"_3")`. To do this, multiply the given mass by the inverse of the molar mass (mol/g).

https://www.ncbi.nlm.nih.gov/pccompound?term=ozone

`19.4color(red)cancel(color(black)("g O"_3))xx(1"mol O"_3)/(47.997color(red)cancel(color(black)("g O"_3)))="0.404 mol O"_3` to three significant figures

Gas Constant, `R`

The gas constant is determined by the units used in the question. The pressure is in mmHg, the volume is in liters (L), the quantity is in moles (n), and the temperature is in Kelvins (K).

`R=62.363577 L*mmHg*K^(-1)*mol^(-1)`

https://en.wikipedia.org/wiki/Gas_constant (1 Torr=1 mmHg)

Organize the data:

Known

`P="1820 mmHg"`

`V="5.12 L"`

`n="0.404 mol"`

`R=62.363577 L*mmHg*K^(-1)*mol^(-1)`

Unknown

`T`

Solution

Rearrange the equation to isolate `T`. Insert the data into the new equation and solve.

`T=(PV)/(nR)`

`T=((1820color(red)cancel(color(black)("mmHg")))xx(5.12color(red)cancel(color(black)("L"))))/((0.404color(red)cancel(color(black)("mol")))xx(62.363577 color(red)cancel(color(black)(L))*color(red)cancel(color(black)(mmHg))*K^(-1)*color(red)cancel(color(black)(mol^(-1))))="370. K"` or `3.70xx10^2 "K"` (rounded to three significant figures)

Convert Temperature from Kelvins to degrees Celsius

Subtract `273.15` from `370^@"C"`.

`"370 K"-"273.15"=97^@"C"` (rounded to a whole number)