A toy balloon originally held 1.00 g of helium gas and had a radius of 10.0 cm.

During the night, 0.25 g of the gas effused from the balloon. Assuming ideal
gas behavior under these constant pressure and temperature conditions, what
was the radius of the balloon the next morning?

The balloon had 1.00g He initially and a radius of 10.0 cm. Calculate the volume of the sphere (V = (4/3)*pi*r^3) and convert 1.00 g He to mols. Use PV = nRT and solve for P (no T is given so assume a convenient number--then use that in all other calculations).

Then 0.25 g He is removed which leaves 0.75g. Convert to mols, use PV = nRT and solve for the new V (use P from the first calculation and T you assumed). Convert V to the new radius. Don't forget that V in PV = nRT is in L but if you use cm for the radius and (4/3)*pi*r^, that V is in cc.

To solve this problem, we can use Boyle's Law and the equation for the effusion rate of a gas.

1. Boyle's Law states that for a fixed amount of gas at constant temperature, the product of pressure and volume is constant. Mathematically, it can be written as:

P1 * V1 = P2 * V2

Where P1 and P2 are the initial and final pressures, and V1 and V2 are the initial and final volumes.

2. In this problem, the pressure is constant, and we need to find the change in volume. The volume of a balloon is directly proportional to its radius cubed, so:

V1 = (4/3)π * R1^3
V2 = (4/3)π * R2^3

Where R1 and R2 are the initial and final radii, respectively.

3. The effusion rate of a gas can be calculated using Graham's Law, which states that the ratio of the effusion rates of two gases is equal to the square root of the inverse ratio of their molar masses. Mathematically, it can be written as:

Rate1 / Rate2 = sqrt(M2 / M1)

Where Rate1 and Rate2 are the effusion rates of the two gases, and M1 and M2 are their molar masses.

4. In this problem, we know the initial and final masses of the gas and want to find the change in volume. The mass of a gas is directly proportional to its molar mass and volume, so we can write:

Rate1 / Rate2 = sqrt((MGas / MHe) * (VHe / VGas))

Where MGas and MHe are the molar masses of the gas and helium, and VHe and VGas are the initial and final volumes of helium and the gas, respectively.

Now let's solve the problem:

Given:
Mass of helium initially (mHe_initial) = 1.00 g
Mass of helium finally (mHe_final) = 0.75 g
Initial radius (R1) = 10.0 cm

1. Calculate the initial volume (V1) using the initial radius:
V1 = (4/3)π * R1^3

2. Use the effusion rate equation to find the ratio of the effusion rates:
Rate1 / Rate2 = sqrt((MGas / MHe) * (VHe / VGas))

3. Since the effusion rate is directly proportional to the mass, we can write:
Rate1 / Rate2 = (mHe_final / mHe_initial)

Rearrange the equation to solve for VHe / VGas:
(MGas / MHe) * (VHe / VGas) = (mHe_final / mHe_initial)^2

4. Substitute the known values to find VHe / VGas:
(VHe / VGas) = (mHe_final / mHe_initial)^2 / (MGas / MHe)

5. Calculate the final volume (VGas) using the equation from step 1 and the ratio VHe / VGas:
VGas = V1 * (VHe / VGas)

6. Finally, calculate the final radius (R2) using the final volume:
R2 = ((3 * VGas) / (4π))^0.33

By following these steps, you should be able to calculate the final radius of the balloon the next morning.

9.08cm

We need the solution not the explanation