The explanation for which the Van der Waals equation reduced to ideal gas law when density is reduced and the reason for which the pressure and volume would no longer need to be modified in Van der Waals equation.

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A)An ideal gas is confined to a container at a temperature of 330 K.What is the average kinetic energy of an atom of the gas? (Express your answer to two significant figures.) B)2.00 mol of the helium is confined to a 2.00-L container at a pressure of 11.0 atm. The atomic mass of helium is 4.00 u, and the conversion between u and kg is 1 u = 1.661 ××10−27 kg.Calculate vrmsvrms. (Express your answer to three significant figures.) C)A gold (coefficient of linear expansion α=14×10−6K−1α=14×10−6K−1 ) pin is exactly 4.00 cm long when its temperature is 180∘∘C. Find the decrease in long of the pin when it cools to 28.0∘∘C? (Express your answer to two significant figures.)

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Chapter 20, Problem 44PQ

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The explanation for which the Van der Waals equation reduced to ideal gas law when density is reduced and the reason for which the pressure and volume would no longer need to be modified in Van der Waals equation.

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A)An ideal gas is confined to a container at a temperature of 330 K.What is the average kinetic energy of an atom of the gas? (Express your answer to two significant figures.) B)2.00 mol of the helium is confined to a 2.00-L container at a pressure of 11.0 atm. The atomic mass of helium is 4.00 u, and the conversion between u and kg is 1 u = 1.661 ××10−27 kg.Calculate vrmsvrms. (Express your answer to three significant figures.) C)A gold (coefficient of linear expansion α=14×10−6K−1α=14×10−6K−1 ) pin is exactly 4.00 cm long when its temperature is 180∘∘C. Find the decrease in long of the pin when it cools to 28.0∘∘C? (Express your answer to two significant figures.)

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The mean free path λ and the mean collision time T of molecules of a diatomic gas with molecular mass 6.00 x10^-25 kg and radius r=1.0x10^-10m are measured.From these microscopic data we can obtain macroscopic properties such as temperature T and pressure P? If yes, consider λ=4.32x10^-8m and T=3.00x10^-10s and calculate T and P.a)It's not possible.b)Yes,T=150K and P~2.04atm.c)Yes,T=150K and P~4.08atm.d)Yes,T=300K and P~4.08atm.e)Yes,T=300K and P~5.32atmf)Yes,T=400K and P~4.08atmg)Yes,T=400K and P~5.32atm.