# A plastic bag is filled with nitrogen at atmospheric pressure and 22.0 oC. Assume that the mass of the bag is negligible and take the temperature of the outside air to be 0 oC. With the help of this "balloon" we want to lift a 50.0 kg girl off her feet.

How large a volume of nitrogen (in m3) is required?

If hot air is used instead of nitrogen, what is the required volume of the balloon if the air inside can be maintained at 37.0 oC ?

## To determine the volume of nitrogen or hot air required to lift the girl, we can use the ideal gas law equation:

PV = nRT

Where:

P = pressure (in this case, atmospheric pressure)

V = volume of the gas

n = number of moles of gas

R = ideal gas constant

T = temperature in Kelvin

First, let's convert the temperatures to Kelvin:

Temperature in Kelvin = Temperature in Celsius + 273.15

Atmospheric pressure is already at standard pressure (1 atm).

Given:

Temperature of nitrogen (T1) = 22.0 oC = 22.0 + 273.15 = 295.15 K

Temperature of hot air (T2) = 37.0 oC = 37.0 + 273.15 = 310.15 K

Assuming the bag is at atmospheric pressure, we can rearrange the ideal gas law equation for volume:

V = (nRT) / P

For nitrogen:

Using the ideal gas constant R = 0.08206 L atm K^-1 mol^-1, which converts to 8.314 J K^-1 mol^-1.

We need to calculate the number of moles of nitrogen required to lift the girl:

n = mass / molar mass

Given:

mass = 50.0 kg

molar mass of nitrogen (N2) = 28.0134 g/mol

n = 50 kg / (28.0134 g/mol)= 1783.05 mol

Substituting the values into the equation for V:

V1 = (n1 x R x T1) / P

V1 = (1783.05 mol x 8.314 J K^-1 mol^-1 x 295.15 K) / (1 atm)

V1 = 481240.329 J / (101.325 J L^-1)

V1 = 4753.14 L

The volume of nitrogen required is approximately 4753.14 liters (L).

For hot air:

Using the same ideal gas constant R = 0.08206 L atm K^-1 mol^-1

Again, we need to calculate the number of moles of air required to lift the girl:

n = mass / molar mass

Given:

mass = 50.0 kg

molar mass of air (approximately) = 28.97 g/mol

n = 50 kg / (28.97 g/mol) = 1724.38 mol

Substituting the values into the equation for V:

V2 = (n2 x R x T2) / P

V2 = (1724.38 mol x 8.314 J K^-1 mol^-1 x 310.15 K) / (1 atm)

V2 = 431017.474 J / (101.325 J L^-1)

V2 = 4255.12 L

The required volume of the balloon if hot air is used is approximately 4255.12 liters (L).