21世纪大学物理(上)答案第七章修正

1、解：已知 ti=15.0C Pi=4.0 105pa t2=37.0C由，查理定律p/T二C3可知Ti=273.15+15.0=288.15KT2=273.15+37.0=310.15KPl/tl=C=P2/t25解的 P2=4.3110 Pa7-255T2=270K解：已知 Pi=1.013 10 Pa P2=3.039 10 Pa Ti=630KPV/T= nKP1V/T1二PV3/TP2V/T2 = PV4TV3/V4= P订2/ P2T1=1/7L1/L2二L1S/L2S二 V3/V4=1/77 3解；P=nKT“7T=P/(nK)=4 nPMHRs/(3MsK)=1.16 10 K7

2、 4P二nKTn 二P/(KT)=2.44 1025p =M/V=PM m°l/kT=1.30Kg m 3 £ =3KT/2=6.21 1021Jd=- =3.45< 1m n7 5PV二MRT/M moiT=PV M m°i/(MR)=90/8.31K_21& =3KT/2=2.24 10J7 6& =3KT/2£ 1= 3KT1/2= 5.5610刃_21£ 2=3KT2/2=7.7210_7 715£ =3KT/2=2.07 10 J2£ =3KT/2=mv212V = (3kT ) =1.574

3、 10 m/sM7-8£ =3KT/2只与T有关所以平动动能相等为£ =3KT=6.21 1J1| 2 v2= (3k)vh=1934MVo'483Vhg s=193O >66o"0 二)V ( (L+0+E+寸+9+"+磐0>S.OH(L+Z+E+寸+9+e+0)、tAZ.-KA9R-KA9>ie-KA 寸次寸空"乂9空0 爲古>衣0|>0 LL >19.6 卜 £>(>lN9)fl0H-L 0X98|<"0 匚 MN9 )、(0 匚 MCOII出-£

4、 o;90ooud 寸9.11 >)m0H>AL>lNHdfro寸卜 匚 >1£>coOL卜寸匚#cod寸匚#9 0匚#山土山d工山H V¥=0击寸E9=o山0E9苗最概然速度为3VO7-14£ =3KT/2只与T有关£ (N2)=6.21 疔 JT=2 £ /(3k)=300K715(2RT/ mmoi)=422m/sf(x)=4 n (m一2冗kTdn/N= £ f(x)dvVp=rkT32 _mV 2)e2kT v dv=4n (丄)n2_ mv2 kTe2V dv =0.83%716图略0 f(v

5、)dv=N f(v)=K kdv=N0kVo= Nk=N/V oVj vf(v)dv二 V O2 2v2= / v f(v)dv=v2/3V。v2 = V°3717(1) 由于分子所允许的速度02V的范围内，由归一的条件可知曲线下的积分为S=Nf(x)dv=N(2) : Nf( x) dv=Nv o2v。av/v odv+adv=N0v oa=2N/3v。3v。vo/(Mmoig)=2858.6m(3) N= vo av/v odv+ 2 adv=7N/12 vo &:2vo=mv=m(3avdv+ 2voV。vo2av dv = N2=31mv367182Z=kTln 卫°P719Z=7t2d nv=2 冗6=3.18 10 Hz7-20P二nkT173n二pv/kT=3.2110 m1入二2 =7.8m2书n7 21Pv/kT=nZ=2d nv8=5.42 107-222d nv8kT =7.1n mkT2冗d p因为p、t