2. A 50.0-kg circular plate of diameter 10.0 cm floatson 0.120 mol of compressed

Question

2. A 50.0-kg circular plate of diameter 10.0 cm floatson 0.120 mol of compressed air, contained in a cylindrical piston of the same diameter with height h. Think of the assembly sitting on a table, in which the plate is above the gas. (a) Calculate the value of h if the temperature is30.0?C. (b) Calculate the distance that the plate moves up ifthe temperature is increased to 100.0?C. (c) Estimate the average distances between air molecules forthe conditions of both parts (a) and (b), assuming the the molecules are uniformlyspread out in the piston. Compare these distances to the size of a diatomicnitrogen molecule (80% of air), approximately 150 pm. 2. A 50.0-kg circular plate of diameter 10.0 cm floatson 0.120 mol of compressed air, contained in a cylindrical piston of the same diameter with height h. Think of the assembly sitting on a table, in which the plate is above the gas. (a) Calculate the value of h if the temperature is30.0?C. (b) Calculate the distance that the plate moves up ifthe temperature is increased to 100.0?C. (c) Estimate the average distances between air molecules forthe conditions of both parts (a) and (b), assuming the the molecules are uniformlyspread out in the piston. Compare these distances to the size of a diatomicnitrogen molecule (80% of air), approximately 150 pm.

Explanation / Answer

Let P is the pressure that the plate cause on gas. P=mg/S=6,24*10^4(Pa) We have PV=nRT so mg/S *S*h=nRT=mgh h=nRT/mg=0,62(m). h=T*nR/mg=0,2(m) Let N is the number of molecule N=Na*n. Assume that each molecule is sphere radius r so v=V/N=4/3*r^3 so v1=6,74*10^-26 so r1=2,52*10^-9(m) v2=8,9*10^-26 so r2=2,77*10^-9(m) Both the distance ˜2,5nm is pretty large compare to 0,15nm ofthe size of a nitrogen molecule

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