ATOMIC - Xe DIATOMIC - N 2 For your assigned gases (atomic, diatomic and molecul

Question

ATOMIC - Xe

DIATOMIC - N2

For your assigned gases (atomic, diatomic and molecular), compute the pressure at T_r = T/T_c = 0.500, T_r = 0.800, T_r = 1.00, T_r = 1.50 and T_r = 2.00 for densities of rho_r = rho/rho_c = 0.500, rho_r = 1.00, rho_r = 1.50 for each of the following equations of state: perfect gas equation; van der Waals equation, or P = RT / V_m - b - a/ V_m^2 where V_m,c = 36, P_c = a/(27b^2) and T_c = 8a/(27b/R) with R being the perfect gas constant; Redlich-Kwong equation, or where V_m,c = 3.84735, P_c = 0.029894A^2/3 R^1/3 B^5/3; and T_c = 0M504[A/(BR)]^2/3: Peng-Robinson equation, or P = RT / V_m - beta - alpha / V_m(V_m + beta) + beta(V_m - beta) where alpha = 0.457236 chi R^2T_c^2/P_c and beta = 0.0777961 RT_C/PC with chi = [1 + (0.37464+ 1.54226 omega - 0.26992 omega^2)(1 - square root T_r)]^2 and with omega being the acentric factor;

Explanation / Answer

(a). For Xe

Atomic weight of xe = 131.29 g/mol

d = P x MW / (R x T)

At Tr = 0.500

0.500 = Pr * 131.29 / (0.0821 * 0.500)

0.0205 = Pr * 131.29

Pr = 0.000156 atm

At Tr = 0.800

0.500 = Pr * 131.29 / (0.0821 * 0.800)

Pr = 0.000250 atm

At Tr = 1.00

0.500 = Pr * 131.29 / (0.0821 * 1.00)

Pr = 0.000312 atm

At Tr = 1.50

0.500 = Pr * 131.29 / (0.0821 * 1.50)

Pr = 0.00047 atm

For N2 -

MW of N2 = 28 g/mol

At Tr = 0.500

1.00 = Pr * 28 / (0.0821 * 0.500)

Pr = 0.00146

At Tr = 0.800

1.00 = Pr * 28 / (0.0821 * 0.800)

Pr = 0.00234

At Tr = 1.00

1.00 = Pr * 28 / (0.0821 * 1.00)

Pr = 0.00293

At Tr = 1.50

1.00 = Pr * 28 / (0.0821 * 1.50)

Pr = 0.00439

For N2O -

MW of N2O = 44 g/mol

At Tr = 0.500

1.50 = Pr * 44 / (0.0821 * 0.500)

Pr = 0.00139

At Tr = 0.800

1.50 = Pr * 44 / (0.0821 * 0.800)

Pr = 0.00268

At Tr = 1.00

1.50 = Pr * 44 / (0.0821 * 1.00)

Pr = 0.00280

At Tr = 1.50

1.50 = Pr * 44 / (0.0821 * 1.50)

Pr = 0.00419

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