Assume the absolute zero apparatus is initially open to the air. The volume of a
ID: 1402035 • Letter: A
Question
Assume the absolute zero apparatus is initially open to the air.
The volume of a sphere is: V = 4r3/3
a) Assume that air pressure is Po = 1.00 atm and the air temperature is To = 11.9o C. How many moles of air will the sphere hold?
nair = moles
b) Suppose now that the apparatus, initially open, is now sealed with the air inside. If the apparatus is now heated to to 888o C, find the pressure inside the apparatus now.
P = atm
c) Assume now that, while maintaining a temperature of 888o, the sphere is injected with another 0.0127 moles of air. What is the new pressure inside the ball?
P = atm
Explanation / Answer
a] Using the ideal gas equation
PV= nRT
we get, n= PV/RT= P0* 4/3*pie*r3 / R T0
n=1*1.013*105 pa*4.187*r3m3/ 8.314* 11.9oC
n=0.043*r3 [ here radius is not specified so n in terms of r]
b] Suppose now that the apparatus is sealed, and the temperature rises to 888o C.
P0/T0 = P1/T1 ...... Amonton's law
1/11.9=P1/888
0.084= P1/888
P1= 0.084*888= 74.62 atm
c] PV= nRT------------------ideal gas law
P~nT
Pnew= R[n+n']T1/ V1=8.314* [0.043*r3 + 0.0127]moles*888oC/ 4/3*pie*r3 = 1763 [0.043*r3 + 0.0127] /r3
Pnew= 1763 [0.043*r3 + 0.0127] /r3 in pascal
1atm=1.0135*105 pa
Pnew= 1763 [0.043*r3 + 0.0127] /r3 /1.0135*105= 1739.8*10-5 [0.043*r3 + 0.0127] /r3
Pnew= [74.81*r3 + 22.09] *10-5 /r3 in atm [ here radius is not specified so P new in terms of r]