Question #71e1b

Dec 17, 2015

$\text{0.609 moles}$

Explanation:

All you have to do here is use the ideal gas law equation to find the number of moles of helium that would occupy that volume under those conditions for pressure and temperature.

The ideal gas law equation looks like this

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

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, equal to $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the temperature of the gas

Now, before plugging in your values, it's important to make sure that the units given to you for volume, pressure, and temperature match the units of the universal gas constant.

In your case, the pressure and volume are expressed in atm and liters, respectively, so no conversion is needed here. However, the temperature is given in degrees Celsius.

Since the universal gas constant uses the temperature in Kelvin, a unit conversion is needed here.

With this in mind, rearrange the ideal as law equation to solve for $n$

$P V = n R T \implies n = \frac{P V}{R T}$

and plug in your values to get

$n = \left(5.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atm"))) * 5.00 color(red)(cancel(color(black)("L"))))/(0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 227.0)color(red)(cancel(color(black)("K}}}}\right)$

$n = \text{0.60883 moles He}$

Rounded to three sig figs, the answer will be

$n = \textcolor{g r e e n}{\text{0.609 moles He}}$