course phy202
Given P=2.1atm, V=8.5m^3, n=1300mol, find v(rms) which is 390m/s according to text. I solve P=1/3 * (N*m*v^2)/V for v=`sqrt(3VP/Nm) and get `sqrt((3*8.5*2.1*1.01E5)/(1300*6.022E23*1300*.014))=1.95E-11m/send of document
Your solution used derived equations.
You can get T from the gas law and KE from kinetic theory, then solve for v. If you do this symbolically you get something close to your equations. Figuring out the quantities at each step you get
T = P V / (n R) = 2*10^5 N/m^2 * 8.5 m^3 / (8.3 J / (mol K) * 1300mol) = 170 K or so.
ave KE is 3/2 k T = 3/2 * 1.38 * 10^-23 J / (particle K) * 170 K = 3.4 * 10^-21 J.
Mass is about 28 amu / molecule * 1.6 * 10^-27 kg / amu = 4 * 10^-26 kg so since 1/2 m v^2 = KE we have
v_rms = sqrt( 2 * KE_ave / m) = sqrt(2 * 3.4 * 10^-21 J / (4 * 10^-26 kg ) ) = 400 m/s^2 or so.
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Good job. Let me know if you have questions.