Question

A pressure cooker is a pot whose lid can be tightly sealed to prevent gas from entering or escaping. Even without knowing how big the pressure cooker is, or what altitude it is being used at, we can make predictions about how much force the lid will experience under different conditions. Part A 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 120∘C? Assume that the temperature of the air outside the pressure cooker is 20∘C (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. Express the force in terms of given variables. View Available Hint(s) F120F 120 F_120 = nothing Part B The pressure relief valve on the lid is now opened, allowing hot air to escape until the pressure inside the cooker becomes equal to the outside pressure pa. The pot is then sealed again and removed from the stove. Assume that when the cooker is removed from the stove, the air inside it is still at 120∘C. What is the magnitude of the net force F20 on the lid when the air inside the cooker has cooled back down to 20∘C?

Answer 1

Answer:

a) F₁₂₀ = 1.34 pa A  , b)  F₂₀ = 0.746 pa A

Explanation:

Part. A .    The definition of pressure is

         P = F / A

As the air can approach an ideal gas we can use the ideal gas equation

        P V = n R T

Let's write this equation for two temperatures

       P₁ V = n R T₁

       P₂2 V = n R T₂

       P₁ / P₂ = T₁ / T₂

point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 393/293

       P₂ = 1.34 pa

We calculate the strength

       P₂ = F₁₂₀ / A

       F₁₂₀ = 1.34 pa A

Part B

In this case the data is

Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure

        P₁ / P₂ = T₁ / T₂

       P₂ = P₁ T₂ / T₁

       P₂ = pa 293/393

       P₂ = 0.746 pa

Let's calculate the force (F20), from this point

      F₂₀ / A = 0.746 pa

     F₂₀ = 0.746 pa A