00:01
We have the rayleigh's form equation p is equal to rt divided by v minus b minus a divided by t power 0 .5 b times b plus b.
00:12
The constant a b are determined by using the equations a is equal to 0 .4278 r square tc power 2 .5 divided by pc and b is 0 .0867 r tc and it is divided by pc where this tc and pc are the critical temperature and critical pressure of vapour.
00:34
Now we talk about the vapour ammonia, ammonia vapour liquid equilibrium is at 120 degree celsius the vapour pressure at 120 degree celsius is 273 plus 120 which is 393 kelvin.
00:48
So is what it is p sat it is 91 .12 bar.
00:53
Now we need to calculate the molar volume and compressibility factor of ammonia vapour at 120 degree celsius.
00:59
The critical temperature in critical preparation of critical pressure of ammonia vapour are first of all critical temperature is 39 is 405 it is 405 .5 and the critical pressure is 112 .8 bar.
01:21
So here we see that ammonia vapour is at t is equal to 393 kelvin and p sat is 91 .12 bar.
01:31
So now we calculate so the constant if we see that we have a and b now we know the value of r into the formula of this a and b this r is what it is gas constant 8 .314 joules per mole kelvin and we have the value of tc as well and we know the value of pc.
01:51
So pc can be converted into 112 .8 multiplied by 10 power 5 n per m square.
01:58
Now we substitute the values of parameters so on substituting the value of a will come out to be equal to 8 .68 n m power 4 k power 0 .5 per mole square...