pv=nrt这个公式是怎么推导出来的？

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)** 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 = nrtwhere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 jpressure ** in kilopascals(kpa)在kilopascals压力(kpa)volume ** in litres(l)音量在升(l)temperature ** 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 atmpressure ** in atmospheres(atm)在大气压力(atm)volume ** in litres(l)音量在升(l)temperature ** in温度在kelvin开尔文kelvin(k)k(k)an ideal gas ** 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 cons**t 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 ** 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 ** 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 beh**iour because:偏离真实气体理想气体行为是因为:at low temperatures the gas molecules h**e 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 beh**iour 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 beh**iour ** 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)÷pwhat volume ** 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 = nrtp = 202.6 kpap = 202.6 kpan = 0.050 moln = 0.050组分的t = 400kt = 400 kv = ? v = ?llr = 8.314 j k-1 mol-1k-1 mol-1 r = 8.314 j202.6v = 0.050 x 8.314 x 400202.6 v = 0.050×8.314×400202.6 v = 166.28202.6 v = 166.28v = 166.28 ÷ 202.6v = 166.28÷202.6v = 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)÷vwhat 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 = nrtp = ? p = ?kpakp** = 7.5lv = 7.5 ln = mass ÷ mmn =质量÷毫米mass = 20.16g质量= 20.16 gmm(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 kr = 8.314 j k-1 mol-1k-1 mol-1 r = 8.314 jp x 7.5 = 10 x 8.314 x 29**×7.5 = 10×8.314×29** x 7.5 = 24360.02p×7.5 = 24360.02p = 24360.02 ÷ 7.5 = 3248kpap = 24360.02÷7.5 = 3248 kpacalculating moles of gas: n = (pv) ÷ (rt)计算摩尔气体:n =(pv)÷(rt)a 50l cylinder ** 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 = nrtp = 10130.0kpap = 10130.0 kp** = 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 jt = 30oc = 30 + 273 = 303k30 oc t = = 30 + 273 = 303 k10130.0 x 50 = n x 8.314 x 30310130.0 x 50 = n×8.314×303506500 = n x 2519.142506500 = n×2519.142n = 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 = nrtp = 253.25kpap = 253.25 kp** = 250ml = 250 ÷ 1000 = 0.250lv = = 250÷250毫升1000 = 0.250 ln = mass ÷ mmn =质量÷毫米mass = 0.40g质量= 0.40 gmm(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,kt = ? t = ?k凯西253.25 x 0.250 = 0.10 x 8.314 x t253.25×0.250 = 0.10×8.314×t63.3125 = 0.8314 x t63.3125 = 0.8314 x吨t = 63.3125 ÷ 0.8314 = 76.15kt = 63.3125÷0.8314 = 76.15公里 20210311