Consider 1 mole of an ideal gas with constant heat capacity

Consider 1 mole of an ideal gas with constant heat capacity C degree_p = 5/2R that changes state from T_1 = 500K and P-1 = 0.6 MPa to T_2 = 200K and P-2 = 0.1 MPa. (a) A possible way of achieving this change is to first at constant pressure P-1 the temperature is changed to T_2 and the pressure is changed to P_2 at constant temperature T_2. Calculate deltaU, Q, and W for each step and the total change. Assume that each step is carried out reversibly. (b) Another possibility to make these changes is to first at constant temperature T_1 the pressure is changed to P_2 and then the temperature is changed to T_2 at a constant pressure P_2. Calculate deltaU, Q, and W for each step and the total change. Assume that each step is carried out reversibly. (c) Which way is more efficient and why?

Solution

Number of moles n = 1

Specific heat at constant pressure Cp = (5/2)R

                                                             = 2.5 R = 2.5x8.314 J/mol K

Tempratures T1 = 500 K

                     T2 =200 K

Pressures P1=0.6Mpa

                     = 0.6 x10 ⁶ Pa

                P2 =0.1MPa

                     = 0.1 x10 ⁶ Pa

Initial volume V1 = RT1/P1

                         =(8.314 x500)/(0.6 x10 ⁶ )

                         = 6.928 x10 ^-3 m ³

(a). At constant pressure , T2 /T1 = V /V1

From this volume V = V1(T2/T1)

                               = ( 6.928 x10 ^-3)(200/500)

                               = 2.77x10 ^-3 m ³

Work done W = P1(V-V1)

                      = 0.6 x10 ⁶ x(2.77-6.928)x10 ^-3

                      = -2494 J

Heat Q = nCp(T2-T1)

            = 1x2.5x8.314 x(200-500)

            = -6235.5 J

Change in internal energy U = Q-W

                                             = -6235.5 -(-2494)

                                              = -3741.5 J

At constant temprature :

PV = constant.

Or P1V = P2V\'

From this volume V \' = P1V /P2

                                 = (0.6Mpa)(2.77x10 ^-3) /(0.1MPa)

                                 = 16.62x10 ^-3 m ³

Work done W \' = RT2 ln(V\'/V)

                        = 8.314 x200 xln(16.62 / 2.77)

                        = 2979.3 J

Heat Q \' = W \'

              = 2979.3 J

Change in internal energy U \' = 0

Total work done = W + W \'

Total heat = Q + Q \'

Total change in internal energy = U+U \'

(b).Constant temprature :

PV = constant

P1V1 = P2V

From this V = P1V1/P2

                    = (0.6MPa)(6.928x10 ^-3)/(0.1MPa)

                    = 41.568 x10 ^-3 m ³

Work done W = RT1 ln(V/V1)

                       = 8.314x500xln(41.568/6.928)

                       = 7448.3 J

Heat Q = W

             = 7448.3 J

Change in internal energy U = 0

Constant pressure :

T2/T1=V\'/V

From this V \' = V(T2/T1)

                     = 41.568x10 ^-3 x(200/500)

                      = 16.62 x10 ^-3 m ³

Work done W \' = P2(V \' - V )

                        = 0.1 x10 ⁶ x(16.62 -41.568)x10 ^-3

                        = -2494 J

Heat Q \' = nCp(T2-T1)

              = 1x2.5x8.314x(200-500)

              = -6235.5 J

Change in internal energy U \' = Q \' -W \'

                                               = -6235.5 J -(-2494)

                                               = -3741.5 J

Total work = W + W \'

Total heat =Q + Q\'

Total change in internal energy = U + U \'

(c). Efficiency = total work / Heat absorbed