How would you calculate the volume of 3.03 g of H2(g) at a pressure of 560 torr and temp of 139 k?

Nov 1, 2015

$\text{47 L}$

Explanation:

I would use the ideal gas law equation

$\textcolor{b l u e}{P V = n R T}$

and rearrange to solve for the volume of the gas, $V$

$V = \frac{n R T}{P} \text{ }$, where

$n$ - the number of moles of the gas
$R$ - the universal gas constant, usually given in as $0.082 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas, expressed in Kelvin
$P$ - the pressure of the gas

Now, you have everything you need to calculate the volume of the gas except the number of moles.

To find how many moles of gas you have in that sample, use hydrogen's molar mass

3.03color(red)(cancel(color(black)("g"))) * "1 mole H"_2/(1.00794color(red)(cancel(color(black)("g")))) = "3.006 moles H"_2

Now plug in your values and solve for $V$ - do not forget to convert the pressure from torr to atm

V = (3.006color(red)(cancel(color(black)("moles"))) * 0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 139color(red)(cancel(color(black)("K"))))/(560/760color(red)(cancel(color(black)("atm")))) = color(green)("47 L")

The answer is rounded to two sig figs, the number of sig figs you gave for the pressure of the gas.