5-1解：(1)325235m/1044.2)27273(1038.11001.1×=+×××==−kTpn(2)RMTpVμ= 335kg/m30.1)27273(31.810321001.1=+××××===∴−RTpVMμρ(3)mn=ρ kg1031.51044.230.12625−×=×==∴nmρ(4)m1045.31044.21193253−×=×==nd5-2解：设容器的质量为m，即放气前容器中气体质量为M1＝mgG−1，放气后容器中气体质量为mgGM−=22由理想气体状态方程有RTmgGRTMVpμμ−==111(1)RTmgGRTMVpμμ−==222(2)(1)、（2）相减VppGGgRT)()(1221−=−μ得)(1221ppGGgVRT−−=μp3时1212333ppGGgVpRTpVM−−⋅===μρ5-3解：RTMpVμ= MRpVTμ=∴J1089.331.8102109.31011038.12323232225323−−−−×=××××××××===MRpVkkTKμε5-4解：kTNKK23=∑=εε总其中N为总分子数kTVNnkTp== kTpVN=∴J150101023232335=××==⋅∴−pVkTkTpVK＝总ε5-5解：时C0°J1065.52731038.12323212310−−×=×××==kTε时C100°J1072.73731038.123232123100−−×=×××==kTε分子具有1eV平均动能时，气体温度为K1073.71038.13106.123232319×=××××==−−KTkε5-6解：由RTMpVμ=有J105.2100.5100.22522335×=××××===−pViRTiMEμ5-7解：RTMpVHH22μ= RTMpVHeHeμ=214222===∴HeHHeHMMμμ又RTiME2μ= 35222222==⋅⋅=∴HeHHHHeHeHHHeHiiiiMMEEμμ5-8解：(1)由理想气体状态方程有molkg102810013.127331.8100.11025.13525−−−×=×××××===pRTpRTVMρμ,气体是N2或CO(2)J1065.52731038.123232123−−×=×××==kTkε 转动自由度235=−=转iJ1077.32731038.122123−−×=××==∴kTi转转ε(3)nkTp= ,3232352/m1069.22731038.110013.1100.1×=×××××==∴−−kTpn容器单位体积内分子的总平动动能J1052.11065.51069.2'32123×=×××=⋅=−kknεε(4)J1070.127331.8253.023×=×××==RTiMEμ5-9解：麦克斯韦速率分布函数22232)2(4)(vekTmvNNvfkTmv−=ΔΔ=ππ令pvvx=mkTvp2=xexNNxΔ=Δ∴−224π当99.0099.100==−=xvvvvppp时，01.101.1100==+=xvvvvppp时，02.0=Δ∴x在0.99vp~1.01vp之间6.102.0)99.0(42)99.0(2=×=Δ−eNNπ％5-10解：（1）速率分布曲线如图。（2）由归一化条件1d)(0=∫∞vvf1d000==∴∫cvvcv得01vc=(3)22dd)(020000vvcvcvvvvfvv==⋅==∫∫∞5-11解：(1)μRTvp2= μRTv32=由题设有μμ1232RTRT=得2312=TTv0Of(v)v（2）由理想气体状态方程RTMpVμ=RTpVMμρ==∴即ρμpRT=ρμpRTv332==5-12解：（1）由μRTvp2=可知，在相同温度下，...
