IB43 Posted January 15, 2020 Report Share Posted January 15, 2020 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 105 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 I could do part a) and part b i) by use of pV=nRT. And I thought using P=F/A will allow me to get F for part ii) from the pressure value I have gotten in part b i). It does not yield 4kN. Could anyone help me out? Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 15, 2020 Report Share Posted January 15, 2020 When the air in the refrigerator is cooled, the force applied by the room air is greater than the force from the air within the refrigerator. Did you try to take the difference and do P = F/A? Reply Link to post Share on other sites More sharing options...
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