【简答题】![](https://cos-cdn.shuashuati.com/shuashuati-web/2024-0521-0905-00/logo-new-ad743.png)
“水平放置且内径均匀的两端封闭的细玻璃管内,有h0=6cm长的水银柱,水银柱左右两侧气柱A、B的长分别为20cm和40cm,温度均为27℃,压强均为1.0×105Pa.如果在水银柱中点处开一小孔,然后将两边气体同时加热至57℃,已知大气压强p0=1.0×105Pa.则管内最后剩下的水银柱长度为多少?”某同学求解如下:因内外压强相等,两侧气体均做等压变化对于A气体, LA1 T1 = LA2 T2 ,LA2= LA1T2 T1 = 330×20 300 cm=22cm对于B气体, LB1 T1 = LB2 T2 ,LB2= LB1T2 T1 = 330×40 300 cm=44cm则剩下水银柱长度=(LA2+LB2 )-(LA1+LB1)-h0问:你同意上述解法吗?若同意,求出最后水银柱长度;若不同意,则说明理由并求出你认为正确的结果.
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