Please tell me which Gas Law equation to use to figure out this problem...

A helium balloon at 25°C, 1 atm, has a volume of 2.88 L. Additional helium is pumped into the balloon. It now contains 1.20 g of He and its new volume is 7.20 L. How many moles of helium gas did the balloon contain originally?

Let x moles of He=moles of He at 2.88L

1.20g *(1 moles/4.0026g)= moles of He at 7.20L

moles of He/7.20L= x moles of He/2.88L

Solve for x moles of He,

x moles of He=(moles of He/7.20L)*2.88L

Thank you! So the Gas Law I would use would be:

Avogadro’s Law Vi/ni = Vf/nf
I would first need to convert the grams to moles and then pop in the known values and solve for the missing value.
I appreciate your help! I did the calculations based on Avogadro's Law and got the correct answer on my practice test.

To solve this problem, we need to use the ideal gas law equation, which is:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)

In this problem, we are given:
P = 1 atm
V = 2.88 L
T = 25°C = 298 K (converted from Celsius to Kelvin)

We need to find the number of moles of helium gas originally in the balloon. Let's assume that n is the number of moles we're looking for.

Now, we can rearrange the ideal gas law equation to solve for n:

n = PV / RT

Substituting the given values:

n = (1 atm) * (2.88 L) / (0.0821 L·atm/mol·K * 298 K)

Simplifying the equation:

n = 0.1122 mol

Therefore, the balloon originally contained approximately 0.1122 moles of helium gas.