2. A system consists of an ideal gas with Cvm 20.86 J mol K (constant over the t

Question

2. A system consists of an ideal gas with Cvm 20.86 J mol K (constant over the temperatures of interest). The gas is initially at temperature 250.0 K, volume 3.000 L, and pressure 1.500 bar. For each of the following processes, calculate the final values of T. P. V. q, w, AU, and AH Show your work and make a table of the results. The initial state of the system is the same for each part (they aren't done sequentially) (a) Constant volume heating to a final temperature of 450.0 K. (c) Constant pressure heating to a final temperature of 450.0K, with the gas expanding against an external pressure of 1.500 bar (c) Heating to a final temperature of 450.0 K and volume of 3.500 L against an external pressure of 1.000 bar. (d) Reversible isothermal expansion to a final pressure of 1.000 bar (e) Adiabatic expansion against an external pressure of 1.000 bar to a final volume of 4.000 L.

Explanation / Answer

Given,

Ideal gas intial conditions,

Cv,m = 20.86 J.mol-1.K-1

P = 1.5 bar

T = 250 K

V = 3.0 L

(a) constant volume to T = 450 K

V(final) = 3 L

T(final) = 450 K

P(final) = nRT(final)/V(final) = 1 x 8.3144 x 450/3 x 10^-3 = 12.47 bar

w = 0 since dV = 0

dU = q = n(Cv,m)dT [n = 1] Feed given values for Cv,m and dT

      = 1 x 20.86 x (450-250) = 4172 J

dH = n(Cp,m)dT [For ideal gas, Cp,m = 5/2 x R with R being gas constant]

      = 1 x 5/2 x 8.3144 x 200 = 4157.24 J

(b) constant pressure heated to final T = 450 K with external pressure for gas expansion 1.5 bar

1.5 bar = 1.5 x 10^5 Pa

P(final) = 1.5 bar

T(final) = 450 K

V(initial) = 3 L = 0.003 m^3

V(final) = (450/250) x 3 x 10^-3 = 5.4 x 10^-3 m^3

w = -P(external)dV = -1.5 x 10^5 x (5.4 x 10^-3 - 3.00 x 10^-3) = -360 J

dU = n(Cv,m)dT = 1 x 20.86 x 200 = 4172 J

q = dH = dU - w = 4172 + 360 = 4532 J

(c) V(final) = 3.5 L = 3.5 x 10^-3 m^3 Pa

T(final) = 450 K

P(final) = nRT/V = 1 x 8.3144 x 450/3.5 x 10^-3 = 10.69 bar

w = -P(ext)dV = -1 x 10^5 x (3.5 x 10^-3 - 3.0 x 10^-3) = -50 J

dU = nCv,mdT = 1 x 20.86 x 200 = 4172 J

q = dH = dU - w = 4172 + 50 = 4222 J

(d) Isothermal expansion

T(final) = 250 K

P(final) = 1 bar = 1 x 10^5 Pa

V(final) = P(intial)V(intial)/P(final) = 1.5 x 3/1 = 4.5 bar

w = -q

dU = 0

w = -nRTln(V(final/V(initial) = -1 x 8.3144 x 250 x ln(4.5 x 10^5/1.5 x 10^5) = -2283.6 J

(e) For adiabatic expansion

q = 0

dU = w = -pdV = -1 x (4-3) = -1 J

V(final) = 4.0 L = 4 x 10^-3 Pa

T(final) = 250 K

P(final) = 1 x 8.3144 x 450/4 x 10^-3 = 9.35 bar

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