You have two gas-filled balloons, one containing He and the other containing H2. The H2 balloon is twice the size of the He balloon. The pressure of gas in the H2 balloon is 1 atm, and that in the He balloon is 2 atm. The H2 balloon is outside in the snow (−5 °C), and the He balloon is inside a warm building (23 °C).

(a) Which balloon contains the greater number of molecules?
(b) Which balloon contains the greater mass of gas?

WEll, the warmer balloon is expanded from the cold, so it at the same PV must have lesser moles of gas.

whats a molecules

To determine which balloon contains the greater number of molecules, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Let's start by converting the temperatures to Kelvin:
Inside temperature (He balloon) = 23 °C + 273.15 = 296.15 K
Outside temperature (H2 balloon) = -5 °C + 273.15 = 268.15 K

Given:
Pressure of H2 balloon (PH2) = 1 atm
Pressure of He balloon (PHe) = 2 atm
Volume of H2 balloon (VH2) = 2VHe (twice the size)
Temperature of H2 balloon (TH2) = 268.15 K
Temperature of He balloon (THe) = 296.15 K

(a) To determine which balloon contains the greater number of molecules, we can compare the number of moles (n) using the formula n = PV / RT:

For H2 balloon:
nH2 = PH2 * VH2 / (R * TH2)

For He balloon:
nHe = PHe * VHe / (R * THe)

Since VH2 = 2VHe, we can substitute this into the equation for nH2:

nH2 = PH2 * 2VHe / (R * TH2)

Now, let's compare the two equations for the number of moles:

nH2 = 1 atm * 2VHe / (R * 268.15 K)
nHe = 2 atm * VHe / (R * 296.15 K)

Since the gas constant R is the same for both balloons, we can ignore it in the comparison:

nH2 = 2VHe / 268.15 K
nHe = VHe / 296.15 K

The H2 balloon has a larger number of moles if nH2 > nHe.
Calculating nH2 / nHe:

nH2 / nHe = (2VHe / 268.15 K) / (VHe / 296.15 K)
nH2 / nHe = (2VHe / VHe) * (296.15 K / 268.15 K)
nH2 / nHe = 2 * 1.1036
nH2 / nHe ≈ 2.2072

Since nH2 / nHe > 1, we can conclude that the H2 balloon contains the greater number of molecules.

(b) To determine which balloon contains the greater mass of gas, we need to consider the molar masses of H2 (2 grams/mole) and He (4 grams/mole).

The mass (m) of the gas can be calculated using the formula m = n * M, where n is the number of moles and M is the molar mass.

For H2 balloon:
mH2 = nH2 * MH2

For He balloon:
mHe = nHe * MHe

Substituting the values into the equations:

mH2 = nH2 * 2 g/mol
mHe = nHe * 4 g/mol

Since the ratio of molar masses is 2:4 or 1:2, and nH2 / nHe ≈ 2.2072 from the previous calculation, we can conclude that the mass of the gas in the H2 balloon is greater than that in the He balloon.

To determine which balloon contains the greater number of molecules, we can use the Ideal Gas Law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

Since we are comparing the two balloons at the same pressure and temperature, we can disregard those variables in our comparison.

For the first question, we need to compare the number of moles of gas in each balloon. We know that the pressure of the H2 balloon is 1 atm and the pressure of the He balloon is 2 atm. The H2 balloon is at a lower temperature (-5 °C) compared to the He balloon (23 °C).

To compare the number of molecules, we need to convert the temperatures to Kelvin:
-5 °C = 268 K
23 °C = 296 K

Now, let's use the Ideal Gas Law equation to compare the number of moles:

n = PV / RT

For the H2 balloon:
n(H2) = (1 atm * V(H2)) / (R * 268 K)

For the He balloon:
n(He) = (2 atm * V(He)) / (R * 296 K)

Since the H2 balloon is twice the size of the He balloon (V(H2) = 2 * V(He)), we can substitute that into the equations:

n(H2) = (1 atm * 2V(He)) / (R * 268 K)
n(He) = (2 atm * V(He)) / (R * 296 K)

As we can see, the number of moles is inversely proportional to temperature. So, the balloon with the lower temperature (H2 balloon in the snow) will have a greater number of moles of gas.

For the second question, we need to compare the mass of gas in each balloon. To do this, we need to consider the molar masses of H2 and He gases.

The molar mass of H2 is 2 g/mol, and the molar mass of He is 4 g/mol.

Using the equations we derived earlier for the number of moles, we can calculate the mass of gas in each balloon:

Mass(H2) = n(H2) * M(H2)
Mass(He) = n(He) * M(He)

Substituting n(H2) and n(He) from earlier:

Mass(H2) = [(1 atm * 2V(He)) / (R * 268 K)] * 2 g/mol
Mass(He) = [(2 atm * V(He)) / (R * 296 K)] * 4 g/mol

Again, since the H2 balloon is twice the size of the He balloon (V(H2) = 2 * V(He)), we can substitute that into the equations:

Mass(H2) = [(1 atm * 2V(He)) / (R * 268 K)] * 2 g/mol
Mass(He) = [(2 atm * V(He)) / (R * 296 K)] * 4 g/mol

Simplifying the equations, we can see that the mass of gas is directly proportional to the number of moles. Therefore, the balloon with the greater number of moles of gas (H2 balloon in the snow) will also contain a greater mass of gas.