1、習(xí)題講解之二,1-26.（教科書第一章習(xí)題的第26題） 因抽氣為等溫過程，故玻意耳定律成立，pV =常量。,解法1：設(shè)抽氣機(jī)每旋轉(zhuǎn)1次可抽氣1次，每次能抽出的氣體體積為V，則每分鐘能抽出的氣體體積（即抽氣機(jī),的抽速）S = V，即 V = S/.p0V = p1(V+V )，p1V = p2(V+V )，pN-1V = pN(V+V ).,p0 = p11+(V/V ) = p21+(V/V )2 = = pN1+(V/V )N.N = ln(p0/pN)/ ln1+(V/V),= ln(p0/pN)/ ln1+S/(V).t1 = N/ = (1/)ln(p0 /pN)/ ln1+S/(V)

2、,= (1 min/400)ln(760/ 1.00)/ ln1+20.0/ (4002.00)= (ln760 min)/ (400ln1.025) 0.67159 min 40.3 s.,本題不用 ，亦可近似求解。因S/(V) = 20.0/(4002.00)= 0.025 1，故可利用ln1+S/(V) S/(V)簡化N的計(jì)算。,N = ln(p0/pN)/ ln1+S/(V) ln(p0/pN)/ S/(V) = V ln(p0/pN)/S.,t1 = N/ (1/)Vln(p0/pN)/S = V ln(p0/pN)/S = 2.00ln(760/1.00)/ (20.0 min-1

3、) 0.66333 min39.8 s,解法2：設(shè)抽氣機(jī)每旋轉(zhuǎn) 1 次可抽氣 2 次，每次能抽出氣體的體積為 V，則每分鐘能抽出的氣體的體積（即抽氣機(jī)的抽速）,S = 2V，即V = S/(2). 同理可得N = ln(p0 /pN) / ln1+(V/V) = ln(p0/pN)/ ln1+S/(2V).,t2 = N/(2) = 1/(2)ln(p0/pN)/ ln1+S/(2V) = 1min/(2400) ln(760/1.00)/ ln1+ 20.0/(24002.00) 0.66747 min 40.0 s.,現(xiàn)在不用 ，仍可近似求解。因S/(2V) = 20.0/(24002.0

4、0) = 0.0125 1，故可利用ln1+S/(2V) S/(2V)簡化計(jì)算。,N = ln(p0/pN)/ ln1+S/(2V) ln(p0 /pN)/ S/(2V) = 2V ln(p0/pN)/S.,t2=N/(2) 1/(2) 2V ln(p0/pN)/S = V ln(p0/pN)/S = 2.00ln(760/1.00)/ (20.0 min-1) 0.66333 min 39.8 s.,解法3：pV = (p+p)(V+V) = pV+pV+Vp+pV pV+pV+Vp.Vp pV = pSt.Vdp pdV pSdt.dt (V/S)(dp/p),將dt對(duì)p從p0到pt積分得t3=dt(V/S)ln(p0/pt) =V ln(p0/pN)/S =2.00ln(760/1.00)/ (20.0 min-1) 0.66333 min39.8 s.,討論：t1 t2 t3 39.8 s的誤差比起t1 40.3 s和t2 40.0 s的誤差稍大些。,但求 t1、t2、t3時(shí)可不用 ，所以當(dāng) 為未知量時(shí)，就必須