Nitrogen is maintained at 400 psia and 100f compare the...

Nitrogen is maintained at 400 psia and -100°F. Compare the specific volume of this nitrogen as predicted by (a) the ideal-gas equation of state, (b) the Benedict-Webb-Rubin equation of state, and (c) the compressibility factor.

Transcript

00:01 Now students here we have a nitrogen that have a pressure of 400 psa and it's kept at the temperature of minus 100 fahrenheit so from this we have to calculate the specific volume of the nitrogen from three methods so first method is ideal gas equation ideal gas equation and in second we have to calculate it from the benedict web rubin equation and in c we have to calculate it from compressibility factor so in section first section we have to calculate it from ideal gas equation so here we can write our ideal gas equation as pv equal to nr t that is our equation number one so from this we can write that is our universal gas constant in terms of psa feet cube divided by mole so that can be written as 10 .75 psia feet cube divided by mole so from we can rewrite this equation number one that is we equal to rt by p so from this we can get our volume as 10 .73 into 359 .67 so here we convert the fahrenheit into the kelvin that is 359 .67 divided by 400.

02:32 So that will give you 9 .67 feet cubed divided by lb.

02:39 So this is the pressure volume of the system at the pressure of 400 p.

02:49 So from this we can we calculated our volume from idle gas equation in section b we have to calculate it from them benedict web rubin equation so the equation can be written as b equal to r t divided by v minus b minus a divided by v into v plus b plus b into v minus b minus b this is our equation number two so here the parameters for them nitrogen a can be written as 1 .39 and b is 0 .03913 so rearranging this equation and we will get our volume as b equal to r t divided by p plus b minus a into r b into t b plus t b plus t r divided by t plus t b into b into f b into 1 by 2 so this is our equation number 3 from this we can calculate our volume so here t r and t b are the 126 .2 fahrenheit and t b that is equal to 97 .1 kelvin so from this we can get above volume as v that is equal to specific volume that be equal to 10 .73 into 35 .967 divided by 400 plus 0 .033 13 in 10 minus 1 .390 divided by 10 .73 into 359 .67 into 400 plus 0 .0391913...

Question

Nitrogen is maintained at 400 psia and -100°F. Compare the specific volume of this nitrogen as predicted by (a) the ideal-gas equation of state, (b) the Benedict-Webb-Rubin equation of state, and (c) the compressibility factor.

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