Onc mole of an ideal monutomic gas is taken through a cycle

Onc mole of an ideal monutomic gas is taken through a cycle ABCA shown in the diagram. The process AB is a reversible isothermal expansion. (o) Find the temperatire of the gas at euch of the points a, b and c? (ie. Ts, Ts, and I?) (b) Find the change in inteenal energy, AFiat, in Joules for each step in the cycle. Show your calculations and fill in your results in the table (c) Calculate the work done, #, in Joulcs for cach step in the cycle. Show your calculations and till in your results in your table. In cach case, indicate (in the table) whether work is done by the system or is done on the system. (d) Find the amount of thermal energy transferred, Q, in Joales for cac mu tahle In rach case, ind

Solution

Pressure at point A is P = 5 atm = 5 x1.01 x10 ⁵ Pa

Volume at point A is V = 5 lit

                                 = 5 x10 ^-3 m ³

Number of moles n = 1 mol

Temprature at point A is T = PV/nR      Where R = gas constant = 8.314 J / mol K

                                      =(5x1.01x10 ⁵x5 x10 ^-3 ) /(1x8.314)

                                     = 303.7 K

Temprature at point B is T \' = T = 303.7 K

Volume at point B is V \' = 25 lit = 25 x10 ^-3 m ³

Pressure at point B is P \' = 1 atm = 1.01 x10 ⁵ Pa

Pressure at point C is P \" = 1 atm = 1.01 x10 ⁵ Pa

Volume at point C is V \" = 5 lit = 5 x10 ^-3 m ³

Temprature at point C is T \" =?

At constant presusre , V \" / V \' = T \" / T \'

From this T \" = T \' ( V \" / V \')

                     = 303.7 ( 5 /25)

                     = 60.74 K

(b). Change in internal energy in isothermal exapnsion U = 0

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

                      = 1x8.314 x303.7 x ln(25/5)

                     = 2525 ln5

                    = 4063.83 J

Heat Q = U + W = 4063.83 J

Along BC process:

it is an isobaric process.

Heat Q \' = nCp(T \" - T \')

Where Cp = Specific heat at constant pressure = 2.5 R = 2.5 x 8.314

So, Q \' = 1 x 2.5 x8.314 x (60.74 - 303.7)

           = - 5049.92 J

Work done W \' = P \' (V \" - V \')

                      = 1.01 x10 ⁵ [(5x10 ^-3) -(25x10 ^-3)]

                      = -2020 J

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

                                          = -5049.92 J + 2020 J

                                         = -3029.92 J

Process CA:

Is is an isochoric process.

Work done W \" = 0

Since in this process the change in volume is zero.
Heat Q \" = nCv ( T - T \")

Where Cv = Specific heat at constant volume

                = 1.5 R = 1.5 x8.314 J / mol K

Substitute values you get , Q \" = 1 x 1.5 x8.314 x ( 303.7-60.74)

                                             = 3029.95 J

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

                                           = 3029.95 J