Five moles of a manatomic ideal gas are in a cylindrical con

Five moles of a manatomic ideal gas are in a cylindrical container (fitted with frictionless piston) at a pressure of 1 atm and at a temp of 300K. First the gas expands isothermally and reversibly to three tiumes its inital volume. Next the gas is compressed to its orginal volume in an isobarix quasi-static process. Finally the system is brough back to its inital pressure along an isochorc quasi-static path.

1) Obtain an expression for the heat absorbed by the gas during each of the three stages and calculate the total heat absorbed during the entire cycle in joules.

Solution

Number of moles n = 5 mol

Initial pressure P = 1 atm = 1.01 x10 5 Pa

Initial temprature T = 300 K

Initial volume V = nRT / P

Where R = Gas constant = 8.314 J / mol K

Substitute values you get V = 5(8.314)(300)/(1.01x10 5)

                                         = 0.1234 m ³

Isothermal expansion:

Final volume V \' = 3V

Final pressure P \' = P / 3           Since in isothermal process PV = P \' V \'

                          = (1/3) atm

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

                     = 5 x8.314x300x ln( 3V / V)

                     = 12471 x ln3

                     = 13700.8 J

Heat Q = W

           = 13700.8 J

Isobaric compression :

Pressure P\' =(1/3) atm

                  = (1/3)(1.01x10 ⁵ Pa)

                  = 33666.66 Pa

Final volume V \" = V

In isobaric process, T \" / T \' = V \" / V \'

Where T \' = T = 300 K

So, T \" / 300 = V /(V/3)

                    = 3

              T \" = 900 K

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

                       = 33666.66 (V -(V/3))

                       = 33666.66 (2V/3)

                       = 22444.44 V

                       = 22444.44x0.1234

                       = 2769.6 J

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

Where Cp = Specific heat at constant pressure = (5/2)R = 2.5 x8.314 J / mol K

So, Q \' = 5 x 2.5 x8.314x(900-300)

   = 62355 J

Isochoric process:

Final pressure P\"\' = P

Volume V\"\' = V \" = V

Final temprature T\"\' = T

Heat absorbed Q \" = nCv (T\"\'-T\")      where Cv = Specific heat at constant volume = 1.5 R = 1.5 x8.314 J/molk

                           = 5 x 1.5 x8.314 x(300-900)

                           = - 37413 J

the total heat absorbed during the entire cycle = Q + Q \' + Q \"

                                                                    = 13700.8 J+62355 J -37413 J

                                                                    = 38642.8 J