Use the compressibility factor to solve VZRTP Near the criti

Use the compressibility factor to solve. V=Z(RT/P)

Near the critical temperature and critical pressure of a gas. it will behave quite like an ideal gas except that its density will be higher than predicted by the Ideal Gas Law. Based on what you know of this \"compressibility\" behavior, what will be the volume in m^3l of 8.5 kg of air if it is at a temperature of 139.125 K. and a pressure of 2.262 MPa?

Solution

>> Now, as given that,

P = Pressure = 2.262 MPa

V = Volume = ?

T = Temperature = 139.125 K

Z = Compressibility Ratio

As, mass of Gas, m= 8.5 Kg

and, seeing from table at Pressure 2.26 MPa & Temperature = 139.125 K

Compressibility factor = 0.8217

>> As, for Non-Ideal Gas,

PV = ZmRT

, R = Gas Constant = 0.287

>> Putting all the values in this equation,

=> 2.262*1000*V = 0.8217*8.5*0.287*139.125

Solving,

V = 0.123 m³ .........REQUIRED VOLUME......ANSWER.....