Question

What volume will 454 grams (1 lb) of hydrogen occupy at 1.05 atm and 25 `"^o`C?

Answer 1

`V = 5.24 xx 10^3` `"L"`

Explanation:

We're asked to find the volume occupied by `454` `"g"` of `"H"_2` at `1.05` `"atm"` and `25` `""^"o""C"`.

To do this, we can use the ideal-gas equation:

`pV = nRT`

where

• `p` is the pressure, in units of `"atm"` (given as `1.05` `"atm"`)

• `V` is the volume the gas occupies, in units of `"L"` (we'll be finding this)

• `n` is the number of moles of the gas present (we'll need to convert the given mass in grams to moles)

• `R` is the universal gas constant, equal to `0.082057("L"·"atm")/("mol"·"K")`

• `T` is the absolute temperature of the gas, in units of `"K"` (given `25` `""^"o""C"`)

We need to do some conversions:

Let's find the number of moles of `"H"_2` present via the molar mass of `"H"_2` (`2.02` `"g/mol"`):

`454cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 225` `"mol H"_2`

The temperature in `"K"` is

`25^"o""C" + 273 = 298` `"K"`

Plugging in known values, and solving for the volume, `V`, we have

`V = (nRT)/p = ((225cancel("mol"))(0.082057("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(298cancel("K")))/((1.05cancel("atm")))`

`= color(red)(5.24xx10^3` `color(red)("L"`