Question 2e2b3

May 15, 2015

The number of moles is $\approx 1.10$ and the mass is $\approx 4.36$ $g$

That's a Chemistry question and you need to use the ideal gas law to find the number of moles $n$.

$p V = n R T$, where $p$ is the pression in Pascal, $V$ the volume in ${m}^{3}$, $R$ is the ideal gas constant ($= 8.134 \left[\frac{J}{K \cdot m o l}\right]$), $T$ the temperature in Kelvin.

Here, we have a ballon with a radius of $r = 18 c m = 0.18 m$.
The volume is given by $V = \frac{4}{3} \pi {r}^{3} \approx 0.0078 \pi$ ${m}^{3}$

The pression $p = 1.05$ $a t m$. We know that $1$ $a t m = 101325$ $P a$.

Therefore, by the Rule of Three :

$p = \frac{1.05 \cdot 101325}{1} = 106391.25$ $P a$

For the temperature, 0° is $273.15$ $K$, so 20° is $293.15$ $K$.

$n = \frac{p V}{R T} = \frac{106391.25 \cdot 0.0078 \pi}{8.134 \cdot 293.15} \approx 1.10$ $m o l$.

For the mass, $m$ $\left[g\right] = n \cdot M$, where M is the molar mass.

The molar mass of $H e$ is $4 \left[\frac{g}{m o l}\right]$.

Therefore, $m \approx 1.10 \cdot 4 \approx 4.36$ $g$.

May 15, 2015

The number of moles of helium will be equal to 1.1 and the mass of helium will be 4.4 g.

SIDE NOTE Small correction - helium cannot exist as a diatomic molecule, so the correct notation is $\text{He}$, not ${\text{He}}_{2}$.

There's nothing wrong with the method used in the other answer, but I want to show you how you can get the number of moles of helium without doing that many conversions.

You that your balloon has a radius of 18 cm. You can determine its volume in ${\text{cm}}^{3}$ by

${V}_{\text{balloon" = 4/3 * pi * "radius}}^{3}$

${V}_{\text{balloon" = 4/3 * pi * ("18 cm")^3 = "24429 cm}}^{3}$

Use the fact that ${\text{1 L" = "1 dm}}^{3}$ to express the volume in liters

24429cancel("cm"^3) * ("1 L")/(10^3cancel("cm"^3)) = "24.429 L"

Now use the ideal gas law equation to solve for $n$ by using $R$ expressed in L atm/mol K

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

n = (1.05cancel("atm") * 24.429cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 20)cancel("K")) = color(green)("1.1 moles")#

To get the mass of helium, use its molar mass, which expresses the mass of 1 mole of helium

$1.1 \cancel{\text{moles He") * "4 g"/(1cancel("mole He")) = color(green)("4.4 g}}$