PV=nRT这个公式是怎么推导出来的？

Answer 1

pV=nRT(克拉伯龙方程）
p为气体压强，单位Pa。V为气体体积，单位m3。n为气体的物质的量，单位mol，T为体系温度，单位K。R为比例系数，数值不同状况下有所不同，单位是J/（mol·K）

推导经验定律
　　（1）玻意耳定律（玻—马定律） 　　当n，T一定时 V，p成反比，即V∝（1/p）①
（2）查理定律 　　当n，V一定时 p，T成正比，即p∝T ②
（3）盖-吕萨克定律 　　当n，p一定时 V，T成正比，即V∝T ③
　　（4）阿伏伽德罗定律 　　当T，p一定时 V，n成正比，即V∝n ④
　　 由①②③④得 　　V∝（nT/p） ⑤
　　 将⑤加上比例系数R得 　　V=（nRT）/p 即pV=nRT

实际气体中的问题当理想气体状态方程运用于实际气体时会有所偏差，因为理想气体的基本假设在实际气体中并不成立。如实验测定1 mol乙炔在20℃、101kPa时，体积为24.1 dm，，而同样在20℃时，在842 kPa下，体积为0.114 dm，，它们相差很多，这是因为，它不是理想气体所致。

<这个高中学过，不好意思，书本上的我记不清了，这个是在百度百科搜的>

Answer 2

推导我不清楚。不过我知道个外国网站有关于这个的一些内容。复制粘贴给你会不会有帮助？中英文应该好些。
如下:

关键概念
An Ideal Gas (perfect gas)is one which obeys理想气体(理想气体)是一个服从Boyle's Law玻意耳定律and和Charles' Law查尔斯王储的法律exactly.没错。

An Ideal Gas obeys the Ideal Gas Law (General gas equation):遵循理想气体理想气体定律(一般气体方程):

PV = nRTPV = nRT

where在
P=pressureP =压力
V=volume体积V =
n=n =moles痣of gas气体的
T=T =temperature温度
R = gas constant (dependent on the units of pressure, temperature and volume)气体常数R =(依赖于单位的压力、温度和体积)
R = 8.314 J K-1 mol-1 ifK-1 mol-1 R = 8.314 J
Pressure is in kilopascals(kPa)在kilopascals压力(kPa)
Volume is in litres(L)音量在升(L)
Temperature is in温度在kelvin开尔文(K)(K)
R = 0.0821 L atm K-1 mol-1 ifR = 0.0821升K-1 mol-1 atm
Pressure is in atmospheres(atm)在大气压力(atm)
Volume is in litres(L)音量在升(L)
Temperature is in温度在kelvin开尔文Kelvin(K)K(K)

An Ideal Gas is modelled on the理想气体蓝本Kinetic Theory of Gases气体动力学理论which has 4 basic postulates:有四个基本假定:

Gases consist of small particles (molecules) which are in continuous random motion气体由小颗粒(分子)都是在连续随机运动

The volume of the molecules present is negligible compared to the total volume occupied by the gas分子的体积相比,是可以忽略不计的总量占用的气体

Intermolecular forces are negligible力量时可以忽略不计

Pressure is due to the gas molecules colliding with the walls of the container压力是由于气体分子的碰撞对容器壁的

Real Gases deviate from Ideal Gas Behaviour because:偏离真实气体理想气体行为是因为:
at low temperatures the gas molecules have less kinetic energy (move around less) so they do attract each other在较低的温度下的气体分子有更少的动能(移动更少),这样他们就都相互吸引

at high pressures the gas molecules are forced closer together so that the volume of the gas molecules becomes significant compared to the volume the gas occupies在高压气体分子被迫更紧密地联系在一起,气体分子的体积会更大的体积相比,气体占据

Under ordinary conditions, deviations from Ideal Gas behaviour are so slight that they can be neglected一般情况下,偏离理想气体的行为很轻微,他们可以忽略不计

A gas which deviates from Ideal Gas behaviour is called a non-ideal gas.一种气体,它偏离理想气体的行为被称为一个理想的气体。

Ideal Gas Law Calculations理想气体定律计算
Calculating Volume of Ideal Gas: V = (nRT) ÷ P计算体积的理想气体:V =(nRT)÷P
What volume is needed to store 0.050 moles of helium gas at 202.6kPa and 400K?需要多大商店0.050摩尔氦气在202.6和400 K kPa吗?

PV = nRTPV = nRT

P = 202.6 kPaP = 202.6 kPa
n = 0.050 moln = 0.050组分的
T = 400KT = 400 K
V = ? V = ?Ll
R = 8.314 J K-1 mol-1K-1 mol-1 R = 8.314 J

202.6V = 0.050 x 8.314 x 400202.6 V = 0.050×8.314×400
202.6 V = 166.28202.6 V = 166.28
V = 166.28 ÷ 202.6V = 166.28÷202.6
V = 0.821 L (821mL)V = 0.821 L(821毫升)

Calculating Pressure of Ideal Gas: P = (nRT) ÷ V计算理想气体的压力:P =(nRT)÷V
What pressure will be exerted by 20.16g hydrogen gas in a 7.5L cylinder at 20oC?什么压力施加于20.16克氢气7.5 L气缸在20摄氏度吗?

PV = nRTPV = nRT

P = ? P = ?kPakPa
V = 7.5LV = 7.5 L
n = mass ÷ MMn =质量÷毫米
mass = 20.16g质量= 20.16 g
MM(H2) = 2 x 1.008 = 2.016g/mol毫米(H2)= 2×1.008 = 2.016克/组分
n = 20.16 ÷ 2.016 = 10moln = 20.16÷2.016 = 10组分
T = 20o = 20 + 273 = 293KT = 20阿= 20 + 273 = 293 K
R = 8.314 J K-1 mol-1K-1 mol-1 R = 8.314 J

P x 7.5 = 10 x 8.314 x 293P×7.5 = 10×8.314×293
P x 7.5 = 24360.02P×7.5 = 24360.02
P = 24360.02 ÷ 7.5 = 3248kPaP = 24360.02÷7.5 = 3248 kPa

Calculating moles of gas: n = (PV) ÷ (RT)计算摩尔气体:n =(PV)÷(RT)
A 50L cylinder is filled with argon gas to a pressure of 10130.0kPa at 30oC. 一个50 L缸充满氩气以一个压力10130.0 kPa 30度。How many moles of argon gas are in the cylinder?多少摩尔氩气在缸?

PV = nRTPV = nRT

P = 10130.0kPaP = 10130.0 kPa
V = 50LV =每50 L水
n = ? n = ?mol组分的
R = 8.314 J K-1 mol-1K-1 mol-1 R = 8.314 J
T = 30oC = 30 + 273 = 303K30 oC T = = 30 + 273 = 303 K

10130.0 x 50 = n x 8.314 x 30310130.0 x 50 = n×8.314×303
506500 = n x 2519.142506500 = n×2519.142
n = 506500 ÷ 2519.142 = 201.1moln = 506500÷2519.142 = 201.1组分

Calculating gas temperature: T = (PV) ÷ (nR)计算燃气温度:T =(PV)÷(nR)
To what temperature does a 250mL cylinder containing 0.40g helium gas need to be cooled in order for the pressure to be 253.25kPa?到什么温度下缸250毫升含0.40克氦气需要冷却为253.25 kPa压力吗?
PV = nRTPV = nRT

P = 253.25kPaP = 253.25 kPa
V = 250mL = 250 ÷ 1000 = 0.250LV = = 250÷250毫升1000 = 0.250 L
n = mass ÷ MMn =质量÷毫米
mass = 0.40g质量= 0.40 g
MM(He) = 4.003g/mol毫米(他)= 4.003克/水解,
n = 0.40 ÷ 4.003 = 0.10moln = 0.40÷4.003 = 0.10组分
R = 8.314 J K mol-1mol-1 R = 8.314 J,K
T = ? T = ?K凯西

253.25 x 0.250 = 0.10 x 8.314 x T253.25×0.250 = 0.10×8.314×T
63.3125 = 0.8314 x T63.3125 = 0.8314 x吨
T = 63.3125 ÷ 0.8314 = 76.15KT = 63.3125÷0.8314 = 76.15公里

Answer 3

推导过程：
对于同物质的量的气体n=1
有三个方程 PV=C1 P/T=C2 V/T=C3
三个相乘 （PV/T）^2=C1C2C3
所以PV/T=√C1C2C3=C (C为任意常数)
然后取一摩尔任意气体,测出P 、 V、T计算出常数C
例如在摄氏0度，T=273 P=P0 V=22.4L
计算出的数值定为R，然后当n增大后，保持R不变，T不变，则V变为nV，即
PV＇/ T=R PnV＇ =nR T nV＇=V PV=nRT
实际气体中的问题当理想气体状态方程运用于实际气体时会有所偏差，因为理想气体的基本假设在实际气体中并不成立。如实验测定1 mol乙炔在20℃、101kPa时，体积为24.1 dm，，而同样在20℃时，在842 kPa下，体积为0.114 dm，，它们相差很多，这是因为，它不是理想气体所致。

Answer 4

推导过程：
对于同物质的量的气体n=1
有三个方程 PV=C1 P/T=C2 V/T=C3
三个相乘 （PV/T）^2=C1C2C3
所以PV/T=√C1C2C3=C (C为任意常数)
然后取一摩尔任意气体,测出P 、 V、T计算出常数C
例如在摄氏0度，T=273 P=P0 V=22.4L
计算出的数值定为R，然后当n增大后，保持R不变，T不变，则V变为nV，即
PV＇/ T=R PnV＇ =nR T nV＇=V PV=nRT