A cylinder of height H = 1.5 m high rests on a table. The bottom half of the cyl

Question

A cylinder of height H = 1.5 m high rests on a table. The bottom half of the cylinder is filled with an ideal gas; on top of that is a frictionelss, movable disk of negligible mass and thickness. Above the disk, the top half of the cylinder is completely filled with liquid mercury, and the top is open to the air. Initially, the gas is at temperature T1 = 362 K; air pressure is Po = 1.013x105 Pa, and the density of mercury is ?Hg = 13600 kg/m3. Now the temperature of the gas is raised until one-half of the mercury spills out of the cylinder. Assume you can ignore the expansion of the mercury and the cylinder due to change in temperature, and evaporation of mercury is negligible. Find T2, the new temperature of the system.
HINT: What do you know about the pressure in the gas?
8 K

Explanation / Answer

let the initial pressure,volume and temp. of gas be p1,v1 and t1 repectively.

t1=362(given)

now pressure on the gas will be due to air and mercury therefre

p1=P0+gh (=density of murcury,g=acceleration due to gravity,h=hight of mercury column)

p1=1.013x105+13600x9.8x(1.5/2)

p1=2.0126x105

and let v1=v

now for ideal gas,

            PV=nRT

applying for the initial condition

          p1v1=nRt1

       (2.0126x105)v=nR(362)                                                      equation (1)

now, after increasing the temperature let final pressure, volume and temperature be p2,v2,t2.

p2=1.013x105+13600x9.8x(1.5/4)

p2=1.5128x105

and now v2=v+v/2=3v/2;

now applying ideal gas equation

     (1.5128x105)(3v/2)=nRt2                                                    equation (2)

now dividing equation 1 and 2 we will get

t2=408.14K

therefore final telmperature will be 408.14K

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