Answer:
a) the mass of air = 8.24 grams
b) the mass of helium = 1.14 grams
c) the mass difference = 7.10 grams
Explanation:
Step 1: Data given
Volume of the tire = 860 mL
Total pressure = 120 psi
Temperature = 26°C
molar mass of air = 28.8 g/mol
Step 2: Convert psi to atm
(
120 psi) (1 atm / 14.7 psi) = 8.163
Step 4: Calculate moles
PV = nRT
⇒ with P = the pressure = 8.163 atm
⇒ with V = the volume = 860 mL = 0.860 L
⇒ with n = the number of moles = TO BE DETERMINED
⇒ with R = the universal gas constant = 0.08206 L*atm/K*mol
⇒ with T = the temperature = 26 °C = 299 Kelvin
n = (8163*0.860)/(0.08206*299)
n = 0.2861 moles of gas
Step 5: Calculate the mass of air in an air-filled tire.
Mass = moles * molar mass
Mass = (0.2861 moles of gas) (28.8 g/mol)
Mass = 8.24 grams
Step 6: Calculate the mass of helium in a helium- filled tire.
mass of helium = 0.2861 moles of gas * 4 g/mol)
mass of helium = 1.14 grams
Step 7: What is the mass difference between the two?
Δmass= 8.24 grams - 1.14 grams
Δmass= 7.10 grams
increases in solution resulting removal of
by forming
. Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more
will dissolve
is added then concentration of
increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less