Making the scaling transformation as in the previous problems, we find that \(\displaystyle \bar{v^2}=∫^∞_0\frac{4}{\sqrt{π}}(\frac{m}{2k_BT})^{3/2}v^2v^2e^{−mv^2/2k_BT}dv=∫^∞_0\frac{4}{\sqrt{π}}\frac...Making the scaling transformation as in the previous problems, we find that \(\displaystyle \bar{v^2}=∫^∞_0\frac{4}{\sqrt{π}}(\frac{m}{2k_BT})^{3/2}v^2v^2e^{−mv^2/2k_BT}dv=∫^∞_0\frac{4}{\sqrt{π}}\frac{2k_BT}{m}u^4e^{−u^2}du.\) As in the previous problem, we integrate by parts: \(\displaystyle ∫^∞_0u^4e^{−u^2}du=[−\frac{1}{2}u^3e^{−u^2}]^∞_0+\frac{3}{2}∫^∞_0u^2e^{−u^2}du.\) Again, the first term is 0, and we were given in an earlier problem that the integral in the second term equals \(\displays…
Making the scaling transformation as in the previous problems, we find that \(\displaystyle \bar{v^2}=∫^∞_0\frac{4}{\sqrt{π}}(\frac{m}{2k_BT})^{3/2}v^2v^2e^{−mv^2/2k_BT}dv=∫^∞_0\frac{4}{\sqrt{π}}\frac...Making the scaling transformation as in the previous problems, we find that \(\displaystyle \bar{v^2}=∫^∞_0\frac{4}{\sqrt{π}}(\frac{m}{2k_BT})^{3/2}v^2v^2e^{−mv^2/2k_BT}dv=∫^∞_0\frac{4}{\sqrt{π}}\frac{2k_BT}{m}u^4e^{−u^2}du.\) As in the previous problem, we integrate by parts: \(\displaystyle ∫^∞_0u^4e^{−u^2}du=[−\frac{1}{2}u^3e^{−u^2}]^∞_0+\frac{3}{2}∫^∞_0u^2e^{−u^2}du.\) Again, the first term is 0, and we were given in an earlier problem that the integral in the second term equals \(\displays…
