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我们来多做点题吧

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关于理想气体状态方程的题目

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假设我有一罐气体

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当前压强是3个大气压

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假设罐子的体积是

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比如 9升

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现在 气压会如何改变

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如果体积从9升变为3升

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从第一个理想气体状态方程的视频中

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你可以大概感觉到

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如果你有一堆——保持——

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这很重要

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我们保持温度不变

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这很重要 必须明白

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所以在我们的直觉

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在理想气体状态方程背后 我们说

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看 如果我们有一定量的粒子

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以及一定的动能

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他们施加一定的压力

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在容器壁上

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如果我们要让容器变小

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我们有同样数量的粒子

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n不变

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平均动能不变

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所以它们会更频繁地撞上容器壁

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所以当我们使容器缩小时

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当体积变大

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当体积变小时

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气压应该升高

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所以我们看看能不能算出准确数字

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我们可以用理想气体方程了

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压强乘体积等于nRT

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现在 粒子数改变了吗

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在我改变情况缩小了体积时？

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不！

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我们有一样的粒子数

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我只是把容器缩小了

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所以n还是n R不变 那是个常数

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温度不变

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所以我原先的压强乘体积

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等于nRT

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而我的新压强乘体积——

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让我叫这个P1和V1

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然后P2是这个

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不好意思 那是V2

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所以V2是这个 我们要算出P2

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P2是什么？

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我们知道P1乘V1等于nRT

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我们还知道既然气体的温度和

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摩尔数都不变

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P2乘V2等于nRT

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既然它们都等于同一个东西

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我们可以说压强乘体积

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只要温度不变

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就也不变

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所以P1乘V1也等于P2乘V2

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那P1是什么？

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P1 我们原先的压强 是3个大气压

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所以3大气压乘9升等于

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我们的新气压乘3升

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如果我们将等式两边同除以3

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3升就消掉了

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我们还有9大气压

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这有道理

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当你把体积缩小2/3

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或者当你使体积

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变为原来的1/3

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你的压强就会增加3倍

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所以这个乘3 这个乘1/3

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一般这是很有用的知识

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如果温度不变

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那么压强乘体积

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也是一个常数

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现在 你可以更进一步

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如果我们看看PV＝nRT

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我们知道两个量不变

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在我们做的大部分练习中

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就是分子数

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以及显然R不会变

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所以如果我们将两边同除以T

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我们就有了PV/T＝nR

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或者可以说它等于一个常数

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它对于任何系统都是一个常数

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如果我们不改变

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容器中的分子数的话

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所以 如果我们改变压强——

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如果一开始有

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压强1 体积1 以及某个温度1

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那就等于这个常数

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如果我们改变其中一个

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我们回到

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压强2 体积2 温度2

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他们应该还等于这个常数

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所以他们相等

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举例来说 假设我开始有

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1个大气压的压强

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而体积是——

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我换一下单位来试一种不同的做法

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——2立方米

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假设温度是27摄氏度

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我刚写了摄氏度

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以为我想让你永远记住

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你要转换成开氏度

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所以27摄氏度加上273是

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正好300开氏度

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我们的新温度是

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实际上我们来算一下新温度

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会是什么

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假设新压强是2大气压

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压强增加了

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我们让容器缩小

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比如1立方米

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所以容器缩小了一半

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而压强增加了一半

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所以你可以猜一下

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你知道我们让压强升高了

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我来让容器更小一点

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实际上 算了

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我来让压强更大

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让压强变成5个大气压

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现在我们想知道第二个温度是什么

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我们建立起等式

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所以就得到

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2/300大气压乘立方米每开氏度

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等于2/T2 我们的新温度

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然后我们有1500等于2T2

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两边同除以2

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得到了T2等于750开氏度

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这有道理 对吧

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我们让压强升高了这么多

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而同时也减小了体积

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所以温度必须上升

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或者如果这么想

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或许我们升高了温度

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因此也使压强

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变得高多了

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尤其是当我们减小了体积时

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我想最好的理解方法是

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压强升高了这么多

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它增加了五倍

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他从1大气压变为5大气压

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因为一面

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我们将体积缩小了1/2

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所以压强应该加倍了

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我们就到了2大气压

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然后我们让温度高多了

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所以我们也在撞上容器壁

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我们使温度变为750开氏度

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所以温度升高了两倍还多

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于是我们得到了5大气压

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现在 另一件你可能会听到的概念

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是发生什么

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在标准温度和气压下

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我把这些都删了

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标准温度和气压

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我把这些不需要的都删掉

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标准温度和气压

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我提到这个

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是因为虽然它被称作

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标准温度和气压

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有时也称作STP

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世界的一大不幸是

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人们还没有完全标准化

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标准气压和温度是什么

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我在维基百科上查了一下

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你可能会看到的

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在大部分物理课以及标准化考试上的

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标准温度是0摄氏度

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也就是 当然 273开氏度

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标准气压是1个大气压

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在维基百科上

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他们写成101.325千帕斯卡

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或者101,000帕斯卡多一点

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当然 一帕斯卡是一牛顿每平方米

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在所有这些中 单位都是

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最难掌握的部分

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但假如我们设

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这些不同的

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标准温度与压强

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来自于不同的标准设定组织

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他们不能完全达成一致

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但如果我们把这个当成

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标准温度和压强的定义

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所以我们设温度

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等于0摄氏度

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也就等于273开氏度

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压强我们设为1大气压

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也可以写成

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101.325或3/8千帕

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我的问题是如果我有一种理想气体

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在标准温度与压强的情况下

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1升中有多少摩尔？

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不 我换一种说法

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1摩尔会占多少升？

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我再多说一点

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n等于1摩尔

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我想算出体积是什么

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如果我有1摩尔气体

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我有6.02乘10的23次方个这种气体气体的分子

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在标准压强1大气压

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以及标准温度273度

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气体的体积是多少？

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我们应用PV＝nRT

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压强是1大气压

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但记住我们用的是大气压

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1大气压乘体积

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这是我们要解的

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我用紫色做

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等于1摩尔 我们有1摩尔这种气体

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乘R 乘温度 乘273

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现在这是开氏度 这是摩尔

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我们想让体积是升

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所以用哪一个版本的R呢？

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我们用的大气压

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我们想要升的体积

216
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当然 我们有开氏度的摩尔

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所以用这个版本 0.082

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那么这是1

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我们可以忽略这个1

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因此体积等于

221
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0.082乘273开氏度

222
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0.082乘273等于22.4升

223
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所以如果我有任何一种理想气体

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所有的气体都不是完全理想化的理想

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但如果我有一种理想气体

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它在标准温度

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也就是0摄氏度

228
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或者水的熔点

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也就是273开氏度

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我有一摩尔这种气体

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它在标准压强 1大气压

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那个气体应该占22.4升

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如果你想知道多少立方米

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它会占

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你只用把22.4升乘

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现在 这是多少立方米

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对于每一立方米都有1000升

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我知道这看起来很多 但这是真的

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只要想想一立方米有多大

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所以这就是0.0224立方米

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如果你有1大气压的某种东西 一摩尔

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在0摄氏度情况下

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反正这其实是

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很有用的数字

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人们会经常说你有2摩尔

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在标准温度与气压

247
00:12:25,292 --> 00:12:26,966
它会占多少升？

248
00:12:26,966 --> 00:12:29,614
好的 1摩尔占这么多

249
00:12:29,614 --> 00:12:31,780
2摩尔在标准温度与气压

250
00:12:31,780 --> 00:12:33,436
就会占两倍体积

251
00:12:33,436 --> 00:12:34,805
因为你只是用PV＝nRT

252
00:12:34,805 --> 00:12:36,272
然后加倍

253
00:12:36,272 --> 00:12:38,790
其他都不变

254
00:12:38,790 --> 00:12:40,992
压强 其他一切都是固定的

255
00:12:40,992 --> 00:12:43,043
所以如果你把摩尔数加倍

256
00:12:43,043 --> 00:12:44,206
就会把体积加倍

257
00:12:44,206 --> 00:12:46,107
或者如果你把摩尔数减半

258
00:12:46,107 --> 00:12:47,674
就会把它的体积减半

259
00:12:47,674 --> 00:12:49,656
所以一个很有用的知识点是在升的单位中

260
00:12:49,656 --> 00:12:52,015
以及标准温度与气压的情况下

261
00:12:52,015 --> 00:12:52,911
当标准温度与气压

262
00:12:52,911 --> 00:12:56,516
被定义为1大气压以及273开氏度

263
00:12:56,516 --> 00:13:00,159
一种理想气体会占22.4升的体积