作业帮PV=NRT和22.我很困惑一些题比如说if 3.20 grams of a gas(mol .wt.equal to 125) were collected over water at 22摄氏度 and at a barometric pressure of 756 mm of hg,what would be the volume of the gas.我觉得这道题就是用PV=NRT算...
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/11 22:36:05
PV=NRT和22.我很困惑一些题比如说if 3.20 grams of a gas(mol .wt.equal to 125) were collected over water at 22摄氏度 and at a barometric pressure of 756 mm of hg,what would be the volume of the gas.我觉得这道题就是用PV=NRT算...
PV=NRT和22.
我很困惑一些题
比如说
if 3.20 grams of a gas(mol .wt.equal to 125) were collected over water at 22摄氏度 and at a barometric pressure of 756 mm of hg,what would be the volume of the gas.
我觉得这道题就是用PV=NRT算...于是就是(3.2/125*0.0821*(22+273))/(756/760)
但是看答案是
3.2/125*22.4*295/273*760/736
这个为什么要用到22.4和273,标态下的量呢
为什么不能直接用PV=NRT?
PV=NRT和22.我很困惑一些题比如说if 3.20 grams of a gas(mol .wt.equal to 125) were collected over water at 22摄氏度 and at a barometric pressure of 756 mm of hg,what would be the volume of the gas.我觉得这道题就是用PV=NRT算...
我想答案的思路是这样的：
首先,已知 1 摩尔理想气体在标准状态下有 760*22.4=1*R*273.所以,R=760*22.4/273.
将其带入真实气体方程式中
V=nRT/P=(3.2/125)(760*22.4/273)*295/756 = 3.2/125*22.4*(295/273)*(760/756) ===> 即答案.