Complicated Math Problem, solving for V, please help!?

Question

One thousand kg of natural gas at 100 bar (10^5 N/m^2) and 255 K is stored in a tank. If the pressure, p, specific volume, v, and temperature, T, of the gas are related by the following expressed:

p = [(5.18 x 10^-3)T/(v-0.002668)] - (8.91 x 10^-3)/(v^2)

where v is in m^3/kg, T is in K, and p is in bar, determine the volume of the tank in m^3.

Answer 1

100,000 N/m² = [(5.18 × 10ˉ³)(255K)/(v - 0.002668)] - (8.91 × 10ˉ³)/v²

100,000 = (1.3209)/(v - 0.002668) - (8.91 × 10ˉ³)/v²

100,000v²(v - 0.002668) = 1.3209v² - (8.91 × 10ˉ³)(v - 0.002668)

100,000v³ - 266.8v² = 1.3209v² - 8.91 × 10ˉ³v + 0.00002377188

100,000v³ - 268.1209v² + 8.91 × 10ˉ³v - 0.00002377188 = 0

Ick...Not even going to try to factor that monster.

I used the solver here:

http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=equations&s2=solve&s3=basic

v = 0.00268105 m³/kg

So, if I didn't mess up, the actual volume should be:

V = 100kg × 0.00268105 m³/kg = 0.0268105 m³

(That was really ugly, perhaps I messed something up. Please double-check and tell me if you see an error.)

Answer 2

one million. you need to placed the unique equation in slop intercept for, and then remembering that perpendicular strains have opposite reciprical slopes write a sparkling equation with this slope. the basically spectacular cost (b) will substitute, sparkling up for this extensive style by skill of pluging interior the factor you have (2,0) and then placed all of the products mutually to get the respond. 2. V = length x width x height, plug interior the equations simplify then sparkling up. 3. i could use long branch, or synthetic branch, yet i'm not sure in case you found out that yet... 4. plug 2 into the 1st equation and then set that equivalent to -one million and slove for ok 5. returned, i could substitute -one million and a couple of to factors (x-2) (x+one million) and then use synthetic divison.

Answer 3

p = [(5.18 x 10^-3)T/(v-0.002668)] - (8.91 x 10^-3)/(v^2);
10^7Pa =5.18*10^(-3)*255 / (V/1000 -0.002668)] - 8.91*10^(-3)* (1000/V)^2;
10^7 = 1320.9/(V-2.668) –8910/V^2;
10^7*V^2*(V-2.668) -1320.9*V^2 +8910*(V-2.668)=0;
10^7*V^3 –2.668*10^7*V^2 –1320.9V^2 +8910*V –23771.88 =0;
V^3 –2.66813V^2 +0.891*10^(-3)V –2.377188*10^(-3) =0;
2 last terms << 2 first terms; hence V=2.668m^3;
indeed 0.891*10^(-3)*2.688 –2.377188*10^(-3) =0;