Question

How many liters of helium are present in a balloon if the balloon contains 2.23 g of He?
A) 12.5 L He
B) 6.25 L HE
C) 22.4 L He
D) 4.0 L He

Answer 1

1 mol of ideal gas = 22.4 L
Do you know how to convert from grams to moles, then moles to liters?

Answer 2

you need to be given condition specifications (temp, pressure)

Answer 3

True. I was assuming that a balloon would be at standard temp and pressure.

Answer 4

@emily.dossen Were you given the density of the helium?

Answer 5

If you were, let the density be represented by the variable \(\rho\).

Density is mass per volume, right? So \(\rho=\dfrac{m}{V}\).

Answer 6

@theEric that's what I thought too.

Answer 7

Or a grams/mol number, which would be on the internet...

Answer 8

For an ideal gas,
$$
PV=nRT
V=\cfrac{nRT}{P}
$$
http://en.wikipedia.org/wiki/Pv%3Dnrt

Where n is the number of moles, T is Temp in Kelvin and P is pressure in Pascals. R is the Gas Constant: \(R=8.314\,4621~\frac{\mathrm{J}}{\mathrm{mol~K}}\)

http://en.wikipedia.org/wiki/Ideal_gas_constant

There are 4.002602 grams per mole for He.

So

2.3g He is

$$
2.23g \times \cfrac{1~ mole}{4.00260~ g}=0.557 moles
$$



Therefore, Volume is
$$
V=\cfrac{0.577\times 8.314\,4621 \times T}{P}
$$

At 1 atm, P = 101 325 Pa (pascals)
At room temperature, 25\(^\circ C\)=298.2 K

So,
$$
V=\cfrac{0.557\times 8.314\,4621 \times 298.2}{101\,325}\\
\approx 0.0136~m^3
$$

1 cubic meter is 1000 Liters.

So,
$$
V=0.0136\times 1000=13.6~ L
$$

You'll need to adjust this for your given temperature and pressure.

The molar volume of an ideal gas at 1 atmosphere of pressure is

22.414 L/mol at 0 °C
24.465 L/mol at 25 °C

So, at 0 °C

The volume of He is,
$$
22.414 \times 0.557 = 12.5 L
$$

Based on this, it appears that your He gas is probably at 0 °C and 1 atm.

So,

$$
V=\cfrac{0.557\times 8.314\,4621 \times 273.2}{101\,325}\\
\approx 0.01248~m^3\\
\approx 12.5 L
$$