a) using ideal gas law, PV=nRT => n=PV/RT=10^5*0.36/8.31/(273+22)=14.6 mol

The number of molecules is 14.6* 6.02*10^23=8.8*10^24

b)1) same formular, P=(273+5)/(273+22)*10^5=94237 pa

2) Dif(P)=5763Pa. the force on the door is thus 5763*0.72=4.149KN

Dozer R.

asked • 11/01/20The 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

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a) using ideal gas law, PV=nRT => n=PV/RT=10^5*0.36/8.31/(273+22)=14.6 mol

The number of molecules is 14.6* 6.02*10^23=8.8*10^24

b)1) same formular, P=(273+5)/(273+22)*10^5=94237 pa

2) Dif(P)=5763Pa. the force on the door is thus 5763*0.72=4.149KN

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