Hydrogen H 2 (3mol) is in an initial state of 373.15K and0.0813 MPa. The gas is

Question

Hydrogen H2 (3mol) is in an initial state of 373.15K and0.0813 MPa. The gas is then expanded to four times its intialvolume in an isotherma and reversible process. Calculatetemperature, pressure, and volume after the expansion and the workthat is performed by the gas. The same expansion can also beperformed in two steps, an isochoric followed by isobaric step.Calculate the temperature after the first step and the (p,V) work for both steps.

Explanation / Answer

Vf = 4*Vi P = nRT/V W = -integral [P*dV] = -nRT*integral [ dV/V] = -nRT*ln(V) =-nRT*ln(Vf/Vi) = -nRT*ln(4) W = -3 mol*(8.314 J/mol/K)*(373.15 K)*ln(4) = 12902.3904 = 1.29 x104 J Initial Volume   V = nRT/P = 3mol*(8.314 J/mol/K)*373.15K/(0.0831 e6 Pa)= 0.111953863 m3 ~ 112 Liters Final Volume = 4*(112L) = 448 Liters Tfinal = Ti (isothermal) = 373.15 K Pf*Vf = Pi*Vi (isothermal) Pf = Pi*(Vi/Vf) = 0.0813 MPa*(1/4) = 0.020325 MPa ~ 2.03 x10-2 MPa ===================================================== Isochoric step => Constant volume, reducing pressure from0.0813 MPa to 2.03 x 10-2 MPa no work involved, because volume does not change Temperature after isochoric step, Pi/Ti = Pf/Tf Tf = Ti*(Pf/Pi) = 373.15 K *(1/4) = 93.3 K Second step, isobaric = > constant Pressure W = -integral [P*dV] = -P*integral[dV] W = -P*V= -(2.03 e-2 MPa)*(448 L - 112 L)) *(1e6 Pa/1 MPa)*(1 m3/ 1000L) W = 6829.2 J ~ 6.83 x 103 J

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