Question

A 20 cm--diameter vertical cylinder is sealed at the top by a frictionless 22 kg piston. The piston is 79 cm above the bottom when the gas temperature is 304 C.The air above the piston is at 1.00 atm pressure.

a) What is the gas pressure inside the cylinder?
b) What will the pressure be if the temperature is lowered to 20 C?
c) What will the height of the piston be if the temperature is lowered to 20 C?

Answer 1

from the gas equation,

PV = RT ............... (1)

and V = m/ρ ................. (2)

from (1)&(2)

desity ρ = Pm/RT

= (1.013*10^5)(22)/(8.314)(577)

= 464.564 kg/m^3

the pressure P = ρgh

=464.564*9.8*79*10^-2

= 3596.660 N/m^2

= 0.035 Pa

* please post each question separatly...

Answer 2

so pressure=1 atm, temperature=304c+273=577k,
volume = pi *r^2*h=pi*(100)(79)= 7900pi.
pV=nRT solve for n
and use it to find the pressure inside the cylinder in pv=nrt for cylinder.