Question

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping.If an otherwise empty pressure cooker is filled with air of room temperature and then placed on a hot stove, what would be the magnitude of the net force F120 on the lid when the air inside the cooker had been heated to 120C ? Assume that the temperature of the air outside the pressure cooker is 20C(room temperature) and that the area of the pressure cooker lid is A. Take atmospheric pressure to be Pa .Treat the air, both inside and outside the pressure cooker, as an ideal gas obeying PV=NkBT .F120=.............F20=..............

Answer 1

`F_(120)=(135906.9113\ A)\ N`

`F_(20)=(101325\ A)\ N`

Step-by-step explanation:

Given:

• initial temperature of air in cooker,
  `T_i=20+273=293\ K`

• initial pressure of air in the cooker,
  `P_i=101325\ Pa`

∵ it is at normal atmospheric pressure.

• final temperature of air in the cooker,
  `T_f=120+273=393\ K`

Since the cooker has a fixed volume, so this process will be isochoric.

Using Charles' Law:

`(P_i)/(T_i)=(P_f)/(T_f)`

`(101325)/(293)=(P_f)/(393)`

`P_f=135906.9113\ Pa`

Now the force acting at 120°C:

`F_(120)=P_f* A`

`F_(120)=(135906.9113\ A)\ N` where area is in meter square.

and the force acting at 20°C:

`F_(20)=P_i* A`

`F_(20)=(101325\ A)\ N` where area is in meter square.