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.

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

1= 26.17

2= 1.17

3=30.86

4=5.86

just took the assigment on edge

that would be because of both gravity and friction. so     c