题12-3:B 第二草气动理论 8kT 丌nd2, m √2md2 T→>47n,v→>2vo,Z0→>2Z0,0->o dw 题12-4:Bf(0)= Ndv y(odo.dN 题12.8:已知:H2,P=1.013×104,m1=167×103 求解 T m=1.99×10kg,R=6.96×10°m n=ms/(Vmu)=ms/(mu Rs),p=nkT 3 T=4nPm,R3/3km。=1.16×107K第十二章 气体动理论 习 题 课 题12-3：B 2 2 8 1 , 2 , , 2 kT Z nd m nd     v v = = = 0 0 0 0 0 0 0 0 T T v v Z Z → → → → 4 , 2 , 2 ,  题12-4：B ( ) , dN N f n Nd V v = = v ( ) , dN nf d V v v = 题12-8：已知： 14 27 2 30 8 , 1.013 10 , 1.67 10 , 1.99 10 , 6.96 10 , a H S S H P P m kg m kg R m − =  =  求：T=? =  =  解： 4 3 /( ) /( ), 3 S S H S H S n m V m m m R P nkT = = =  3 7 4 / 3 1.16 10 T Pm R km K = =   H S S