Question

5.3 A 6-m3 tank contains helium at 400oK and is evacuated from atmospheric pressure to a pressure of 740 mmHg vacuum. Determine (a) the mass of helium remaining in thetank; (b) the mass of helium pumped out. (c) The temperature of the remaining helium falls to 10oC. What is the pressure in kPa?

Answer 1

Can you use ideal gas law? Or VDW EOS? Second Virial? Correspondingstates? It seems you can just use ideal gas

PV = nRT

P/n = RT/V, because T stays at 400K, and V is 6 m^3

atmospheric pressure: 760 mmHg

P1/n1 = P2/n2
n2 = (P2/P1)*n1

n1 = P1*V1/(R*T1) = 760 mmHg *(1atm/760 mmHg)*(101325 Pa/1atm) * 6m^3 /(8.314 J/mol*K * 400K)
n1 = 182.81 moles

n2 = (740 mmHg/760mmHg) * 182.91 moles = 177.998 moles

remaining mass = 177.998 moles *4.0026 g/mol = 712.45 g

b) del n = n1-n2 = 182.81 - 177.998 = 4.8108 moles

del m = 4.8108 mol *(4.0026 g/mol) = 19.26 grams

c) V = 6 m^3
n = 177.998 moles
T = 10 C = 283 K

PV = nRT
P = nRT/V = (177.998 moles)*(8.314 J/mol*K)*(283K)/6 m3 = 69800.93Pa

69800.93 Pa *(760 mmHg/101325 Pa) = 523.55 mmHg

(It is 10 C, right? If it is 100 C, use 373 K, instead of 283 K. PMme if it 100 C, and I can resolve it.)