题目内容
如图所示,导热性能良好粗细均匀两端封闭的细
玻璃管ABCDEF竖直放置。AB段和CD段装有空气,BC段和DE段为水银,EF段是真空,各段长度相同,即AB=BC=CD=DE=EF,管内AB段空气的压强为p,环境温度为T。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241233412601859.jpg)
(1)若要使DE段水银能碰到管顶F,则环境温度至少需要升高到多少?
(2)若保持环境温度T不变,将管子在竖直面内缓慢地旋转180°使F点在最下面,求此时管内两段空气柱的压强以及最低点F处的压强。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412334124572.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241233412601859.jpg)
(1)若要使DE段水银能碰到管顶F,则环境温度至少需要升高到多少?
(2)若保持环境温度T不变,将管子在竖直面内缓慢地旋转180°使F点在最下面,求此时管内两段空气柱的压强以及最低点F处的压强。
(1)设初状态每段的长度为h,CD段空气柱末状态的长度为hCD
根据等压变化,对CD段空气柱有
(1分)
(1分)
得T1="1.5T " (2分)
(2)设CD段空
气柱末状态的长度为hCD,压强为pCD
根据波意耳定律,对CD段空气柱有
(2分)
对AB段空气柱有
(2分)
得pCD=
(1分)
pAB=
(1分)
pF=
(2分)
根据等压变化,对CD段空气柱有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123341276536.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123341292628.gif)
得T1="1.5T " (2分)
(2)设CD段空
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/2014082412334130765.gif)
根据波意耳定律,对CD段空气柱有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123341323598.gif)
对AB段空气柱有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123341338793.gif)
得pCD=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123341354376.gif)
pAB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123341370320.gif)
pF=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824123341385385.gif)
略
![](http://thumb2018.1010pic.com/images/loading.gif)
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