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

Jul 17, 2017

$V = 5.24 \times {10}^{3}$ $\text{L}$

#### Explanation:

We're asked to find the volume occupied by $454$ $\text{g}$ of ${\text{H}}_{2}$ at $1.05$ $\text{atm}$ and $25$ $\text{^"o""C}$.

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

$p V = n R T$

where

• $p$ is the pressure, in units of $\text{atm}$ (given as $1.05$ $\text{atm}$)

• $V$ is the volume the gas occupies, in units of $\text{L}$ (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 $0.082057 \left(\text{L"·"atm")/("mol"·"K}\right)$

• $T$ is the absolute temperature of the gas, in units of $\text{K}$ (given $25$ $\text{^"o""C}$)

We need to do some conversions:

Let's find the number of moles of ${\text{H}}_{2}$ present via the molar mass of ${\text{H}}_{2}$ ($2.02$ $\text{g/mol}$):

454cancel("g H"_2)((1color(white)(l)"mol H"_2)/(2.02cancel("g H"_2))) = 225 ${\text{mol H}}_{2}$

The temperature in $\text{K}$ is

${25}^{\text{o""C}} + 273 = 298$ $\text{K}$

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

$V = \frac{n R T}{p} = \left(\left(225 \cancel{\text{mol"))(0.082057("L"·cancel("atm"))/(cancel("mol")·cancel("K")))(298cancel("K")))/((1.05cancel("atm}}\right)\right)$

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