Consider a reversible Carnot cycle with 25 mol of an ideal g

Consider a reversible Carnot cycle with 2.5 mol of an ideal gas with Cv= 3/2R as the working substance. The initial isothermal expansion occurs at the hot reservoir temperature of T(hot) = 750. K from an initial volume of 31.1775 L (Va) to a volume of 62.3550 L (Vb). The system then undergoes an adiabatic expansion until the temperature falls to Tcold = 345. K at point c. The system then undergoes an isothermal compression to point d and a subsequent adiabatic compression until the initial state described by Ta = 750 K and Va = 31.1775 is reached.

Calculate w, q, Delta U and Delta H for each step in the cycle and the total cycle. What is the efficiency, epsilon for this heat engine?

Solution

Number of moles n = 2.5

Hot reservoir temprature T = 750 K

Cold reservoir temprature T \' = 345 K

Initial volume V = 31.1775 L

                        = 31.1775 x10 ^-3 m ³

Final volume in isothermal expansion V \' = 62.355 L = 62.355 x10 ^-3 m ³

Isothermal expansion :

Work done W = nRT ln(V \' / V )        Where R = Gas constant = 8.314 J / mol K

                      = 2.5 x 8.314 x750 x ln(62.355/31.1775)

                      = 2.5 x8.314 x 750 xln(2)

                       = 10800 J

Change in internal energy U = 0

Heat Q = U + W

           =10800 J

Adiabatic expansion:

Heat Q \' = 0

Work done W \' = [nR/(r-1)][T-T\']       Where r = ratio of specific heats = 1.67

                       =[(2.5 x8.314)/(1.67 -1)][750-345]

                       = 12564 J

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

                                          = -12564 J

Volume at point C is V \" = ?

In adiabatic process T V ^r-1 = constant

Apply this equation to b and c points,

T V \' ^r-1 = T \' V\" ^r-1

(V\"/V) ^r-1 = T / T \' = 750/345

               = 2.173

V \" / V \' = (2.173) ^1/(r-1)

              = 3.186         Since r = 1.67

         V \" = 3.186 V \'

               = 3.186 x 62.355 x10 ^-3 m ³

               = 0.1987 m ³

Isothermal compression :

We don\'t know the pressure or volume values after isothermal compression.

If you know any one value you find V\"\' i.e., volume of the gas after isothermal compression

Work done W \"= nRT \' ln(V \"\' / V \'\' )        Where R = Gas constant = 8.314 J / mol K

From this you find W \" .