Thermal physics problem -- Pressure and temperature of air in a refrigerator

[ib43]

Homework Statement

I am stuck on one particular section regarding finding force given pressure

Relevant Equations

pv=nRT, P=F/A

There is this one problem from past exam papers which I cannot seem to do:

The air in a kitchen has pressure 1.0 x 10^5 Pa and temperature 22'C. A refrigerator of internal volume 0.36 m^3 is installed in the kitchen.

(a) With the door open the air in the refrigerator is initially at the same temperature and pressure as the air in the kitchen. Calculate the number of molecules of air in the refrigerator.

(b) The refrigerator door is closed. The air in the refrigerator is cooled to 5.0'C and the number of air molecules in the refrigerator stays the same.

(i) Determine the pressure of the air inside the refrigerator.

(ii) The door of the refrigerator has an area of 0.72m^2. Show that the minimum force needed to open the refrigerator door is about 4 kN

My attempts:
a) I simply used the equation pV=NKT where p = 1.0 x 10^5 and T=295, and K = Boltzmann constant, V=0.36. This gave me the number of molecules.
b) i) I used P1/T1=P2/T2 where I am trying to find P2, where P1, T1, are given and T2= 278.
ii) I assumed we can use the pressure found in part b i) then use P=F/A but it does not seem to give me the answer of 4kN.

 

[mjc123]

Why use the pressure from part bi? What is keeping the door shut?

 

[Qfjacobs]

Find the pressure inside using what you got in part B times the area. Then find the pressure outside which is the given pressure times the same area. So, to find the minimum force to open it, you find the difference. So with sig figs taken into consideration, I got 72,000(force outside) - 68,000(force inside) = 4 kN.