Consider a monatomic gas of particles eachwith mass M. What is x, vx,rms = , the

Question

Consider a monatomic gas of particles eachwith mass M. What is x, vx,rms = , the root mean square (rms) of the x component of velocity of the gas particles if the gas is at anabsolute temperature T? Now consider the same system - a monatomic gas of particles of mass M -- except in three dimensions. Find vrans, the rms speed if the gas is at an absolutetemperature T. Express your answer in terms of T, kB, M, and other given quantities. What is the rms speed v0 of molecules in air at 0 degree C? Air is composed mostly of N2 molecules, so you may assume that it has molecules of average atomic mass 28.0 times 1.661 times 10-27 Kg = 4.65 times 10-26 kg Find , the rms angular speed of the dumbbell about asingl axis (taken to b the x axis), assuming that it is in equilibrium at temperature T. What is the typical rotational frequency frot for a molecule like N2 at room temperature (25 degree C)? Assume that d for this molecule is 1A = 10-10 m. Take the atomic mass of N2 to be mN2 = 4.65 times 10-26 kg

Explanation / Answer

rate me .....if A gas molecule has an energy of E = (1/2)·k·T per degree of freedom. where k = 1.38065×10?²³JK?¹ Boltzmann constant T absolute temperature part a In translational motion you have three degrees of freedom (possible motion in x,y and z direction). Hence: E = (3/2)·k·T On the other hand is the translatory kinetic energy of an object of mass m moving with velocity v defined as: E = (1/2)·m·v² Equate the two expression and solve for the speed: (3/2)·k·T = (1/2)·m·v² => v = v(3·k·T/m) = v(3 · 1.38×10?²³JK?¹ · 273.15K / 4.65×10?²6kg) = 493m/s part b A bimolecular gas particle like the nitrogen molecule has two degrees of freedom for rotational motion. E = (2/2)·k·T = k·T The rotational kinetic energy of object of moment of inertia J rotating at angular velocity ? is: E = (1/2)·J·?² Consider the nitrogen molecule as two point masses connected by thin massless rod. Take the middle of the rod as axis of rotation. Moment of inertia equals mass times squared distance to the axis of rotation J = 2 · (m/2)·(d/2)² = (1/4)·m·d² Angular velocity and frequency are related as: ? = 2·p·f Therefore E = (1/2) · (1/4)·m·d² · (2·p·f )² = (1/2)·p²·m·d²·f² and k·T = (1/2)·p²·m·d²·f² => f = v[ 2·k·T/(p²·m·d²) ] = v(2 · 1.38×10?²³JK?¹ · 298.15K / (p² · 4.65×10?²6kg · (10?¹°m)²) ] tahnks

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