Question

Olympic cyclist fill their tires with helium to make them lighter. Calculate the mass of air in an air filled tire and the mass of helium in a helium filled tire. What is the mass difference between the two? Assume that the volume of the tire is 855ml that is filled with a total pressure of 125psi, and that the temperature is 25 degrees Celsius. Also, assume an average molar mass for air of 28.8g/mol

Answer 1

Answer: The mass difference between the two is 7.38 grams.

Step-by-step explanation:

To calculate the number of moles, we use the equation given by ideal gas follows:

`PV=nRT`

where,

P = pressure = 125 psi = 8.50 atm (Conversion factor: 1 atm = 14.7 psi)

V = Volume = 855 mL = 0.855 L (Conversion factor: 1 L = 1000 mL)

T = Temperature =
`25^oC=[25+273]K=298K`

R = Gas constant =
`0.0821\text{ L. atm }mol^(-1)K^(-1)`

n = number of moles = ?

Putting values in above equation, we get:

`8.50atm* 0.855L=n* 0.0821\text{ L atm }mol^(-1)K^(-1)* 298K\\\\n=(8.50* 0.855)/(0.0821* 298)=0.297mol`

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For air:

Moles of air = 0.297 moles

Average molar mass of air = 28.8 g/mol

Putting values in equation 1, we get:

`0.297mol=\frac{\text{Mass of air}}{28.8g/mol}\\\\\text{Mass of air}=(0.297mol* 28.8g/mol)=8.56g`

Mass of air,
`m_1` = 8.56 g

• For helium gas:

Moles of helium = 0.297 moles

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

`0.297mol=\frac{\text{Mass of helium}}{4g/mol}\\\\\text{Mass of helium}=(0.297mol* 4g/mol)=1.18g`

Mass of helium,
`m_2` = 1.18 g

Calculating the mass difference between the two:

`\Delta m=m_1-m_2`

`\Delta m=(8.56-1.18)g=7.38g`

Hence, the mass difference between the two is 7.38 grams.