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Phoenix314
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These problems are from Giancoli 5th edition (principles with applications)
A cubic box of volume 0.039 m^3 is filled with air at atmospheric pressure at 20 celsius. The box is closed and heated to 180 celsius. What is the net force on each side of the box?
I first used the ideal gas law to figure out the pressure after the box is heated. So P1/T1 = P2/T2 because the box is closed and the volume doesn't change, and nR is constant. I then got P2 = 1.56 * 10^5 N/m^2 - but I'm not sure how to use this to figure out the forces on the sides of the box
Calculate approximately the total translational kinetic energy of all the molecules in an E.Coli bacterium of mass 2.0 * 10^-15 kg at 37 celsius. Assume 70% of the cell, by weight, is water, and the other molecules have an average molecular weight on the order of 10^5.
I don't know how to start this problem - Giancoli shows us how to calculate the average KE and how it's proportional to temperature, but not total translational KE. I'm not sure how to use the composition of the E.Coli cells either.
Any help is greatly appreciated! Thank you
A cubic box of volume 0.039 m^3 is filled with air at atmospheric pressure at 20 celsius. The box is closed and heated to 180 celsius. What is the net force on each side of the box?
I first used the ideal gas law to figure out the pressure after the box is heated. So P1/T1 = P2/T2 because the box is closed and the volume doesn't change, and nR is constant. I then got P2 = 1.56 * 10^5 N/m^2 - but I'm not sure how to use this to figure out the forces on the sides of the box
Calculate approximately the total translational kinetic energy of all the molecules in an E.Coli bacterium of mass 2.0 * 10^-15 kg at 37 celsius. Assume 70% of the cell, by weight, is water, and the other molecules have an average molecular weight on the order of 10^5.
I don't know how to start this problem - Giancoli shows us how to calculate the average KE and how it's proportional to temperature, but not total translational KE. I'm not sure how to use the composition of the E.Coli cells either.
Any help is greatly appreciated! Thank you