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5 September, 17:43

A closed, nonconducting, horizontal cylinder is fitted with a nonconducting, frictionless. floating piston which divides the cylinder into Sections A and B. The two sections contain equal masses of air, initially at the same conditions. T_1 = 300 K and P_1 = 1 atm. An electrical heating element in Section A is activated, and the air temperatures slowly increase: T_A in Section A because of heat transfer, and T_B in Section B because of adiabatic compression by the slowly moving piston. Treat air as an ideal gas with C_p = 7/2 R, and let n_A be the number of moles of air in Section A. For the process as described, evaluate one of the following sets of quantities: T_A, T_B, and Q/n_A, if P (final) = 1.25 atm. T_B, Q/n_A, and P (final), if T_A = 425 K. T_A, Q/n_A, and P (final), if T_B = 325 K. T_A, T_B, and P (final), if Q/n_A = 3 kJ mol^-1.

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  1. 5 September, 18:52
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    T_Bf = 319.75 K

    T_Af = 430.25 K

    Q / n_A = 3.118 KJ/mol

    Explanation:

    Given:

    - The initial conditions:

    T_A = T_B = T_i = 300 K

    P_A = P_B = P_i = 1 atm

    - The masses in both sections (n)

    - Cp = 7/2 R

    - R = 8.314 KJ/molK

    Find:

    - For first set a only:

    T_Af, T_Bf, and Q/n_A, if P_f = 1.25 atm

    Solution:

    - Use the ideal gas equation to calculate the individual Volumes of both sections:

    V_A = V_B = n*R*T_i / P_i

    - For initial state the total volume V:

    V = V_A + V_B

    V = n*R*T_i / P_i + n*R*T_i / P_i

    V = 2*n*R*T_i / P_i

    - Similarly the Total Volume V by using final state:

    V = n*R * (T_Af + T_Bf) / P_f

    - Since, the total volume remains same, then equate the initial state and final state total volumes V:

    2*n*R*T_i / P_i = n*R * (T_Af + T_Bf) / P_f

    2*T_i / P_i = (T_Af + T_Bf) / P_f

    - Use, the adiabatic process equation for section B:

    T_Bf = T_Bi * (P_Bi / P_Bf) ^ (1 - k / k)

    Where, k = Cp / Cv and Cv = Cp - R, Cp = 7/2 R

    Cv = 7/2 R - R

    Cv = 5/2 R

    k = (7/2 R) / (5/2 R) = 7/5

    Hence,

    T_Bf = 300 * (1 / 1.25) ^ (-2/7)

    T_Bf = 319.75 K

    And,

    2*T_i / P_i = (T_Af + T_Bf) / P_f

    2*300/1 = (T_Af + 319.75) / 1.25

    T_Af = 600*1.25 - 319.25

    T_Af = 430.25 K

    - The energy transfer by the system is given by first thermodynamic Law:

    Δ U = Q + W

    For no boundary work W = 0,

    Δ U_A + Δ U_B = Q

    Q = n_A*Cv * (T_Af - T_i) + n_B*Cv * (T_Bf - T_i)

    Where, n_A = n_B = n,

    Q / n_A = Cv * (T_Af + T_Bf - 2*T_i)

    Q / n_A = 5*8.314/2 * (319.75 + 430.75 - 2*300)

    Q / n_A = 3.118 KJ/mol
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