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

Concepts and reason

The main concept used to solve this problem is Pressure-Temperature Law.

Initially, find the pressure inside the pressure cooker by using the Pressure-Temperature Law. Then, find the force exerted by the air inside cooker on the lid of the cooker.

Fundamentals

The Pressure-Temperature Law states that the pressure of the of a given gas held at constant volume is directly proportional to the Kelvin temperature. The increase in pressure results in an increase in the temperature and vice-versa.

The expression of the Pressure-Temperature Law is as follows:

Here, are the initial and final pressures and are the initial and final temperatures respectively.

The force in terms of pressure can be given by the following expression.

Here, A is the area on which the pressure is applied and is the change in pressure.

Initially, the gas was at the atmospheric pressure and the temperature of the gas is .

The temperature of the gas initially in terms of Kelvin is as follows:

Substitute for T in the above expression.

Finally, the temperature of the gas is as follows:

Substitute for T in the above expression.

The expression of the Pressure-Temperature Law is as follows:

Rearrange the above expression for .

The difference in the pressure can be written as follows:

Substitute for in the above expression.

Substitute for , 393 K for , and 293 K for in the above expression.

The force on the lid is given as follows:

Substitute for and A for A in the above expression.

The difference in the pressure can be written as follows:

Substitute for in the above expression.

$Av=-=() $

Substitute for , 293 K for , and 393 K for in the above expression.

The force on the lid is given as follows:

Substitute for and A for A in the above expression.

Ans:

The force on the lid is .

The force on the lid is .