Question

Assume the absolute zero apparatus is initially open to the air. The volume of a sphere is: V = 4πr3/3 (radius 2 inch) a) Assume that air pressure is Po = 1.00 atm and the air temperature is To = 15.5o 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 571o C, find the pressure inside the apparatus now. P = atm c) Assume now that, while maintaining a temperature of 571o, the sphere is injected with another 0.0379 moles of air. What is the new pressure inside the ball? P = atm

Answer 1

V = 4πr^3/3 = 4*(22/7)*(2*0.0254)^3/3 = 0.0005493 m^3
T = 15.5⁰C = (15.5+273.15)K = 288.65 K
P = 1.00atm = 1.013x*10^5 Pa

We assume air behavas as an ideal gas.
PV = nRT
n = PV/(RT)
= 1.013*10^5 * 0.0005493 / (8.31 * 288.65)
= 0.023197 mol
______________

571⁰C = (571+273.15)K =844.15K

Since the volume and amount of air are constant, the pressure is proportional to the absolute temperature.

The new pressure = 1.013*10⁵ * (844.15/288.65)
= 2.9625*10⁵ Pa

PV = nRT

P = 0.0379*8.31*844.15/0.0005493 = 4.84*10⁵ Pa