A heat engine carries 4.30 mol of an ideal monatomic gas around the cycle shown

Question

A heat engine carries 4.30 mol of an ideal monatomic gas around the cycle shown in the figure. Process 1-2 takes place at constant volume and takes the gas from 300 K to 600 K, process 2-3 is adiabatic and takes the gas from 600 K to 455 K, and process 3-1 takes place at a constant pressure and returns the gas to 300 K.

Compute the heat Q for the whole cycle.

Compute the change in internal energy DeltaE for the cycle as a whole.

Calculate the work done by the gas for the cycle.

If the initial pressure at point 1 is 2.0 atm, find the pressure at point 2 in atm.

If the initial pressure at point 1 is 2.0 atm, find the volume at point 3 (in m3).

Explanation / Answer

(a)1 to 2 Constant volume and temperature 300K to 600 K

According to PV/T = K    P is proportional to T

So pressure is incresed in this process

(b) 2 to 3 Adiabatic and temperature changes from 600 K to 455K

(c) 3 to 1 constant pressure and tempersture is from 455 K to 300 K

1st law of thermodynamics

u = Q +W

U = Increase in internal energy

Q = Thermal energy supplied to the system

W – work done on the system   W = PV ( Pressure x change of volume)

For process 1 to 2

W = 0 no change of volume

PV/T = constant Pressure proportional to the T

Temperature is doubled and therefore pressure is also doubled

Pressure at point 2 = 2atm

U = Q     

For an ideal monoatomic gas Internal energy = Average kinetic energy = 3/2nRT ( No potential energy as there is no intermolecular force)

U = 3/2 x 4.3 x 8.3 x(600-300) = 16 060.5 J

Thermal energy supplied to the system = 16 060.5 J

For process 2 to 3

Thermal energy supplied = 0 as it is an adiabatic process

For process 3 to 1

The temperature reduced ,so internal energy reduced

U = 3/2 x 4.3 x 8.3 ( 300-455) = - 8297.925 J

PV/T = K   Pressure constant so Volume is proportional to the temperature T

Initial pressure is 2.0 atm     PV = nRT at point 1     1 atm = 1x 105 Pa

V = nRT/P = 4.3 x 8.3x 300/ 2 x 105 = 0.053535m3

Volume at point 3

0.053535/300 = V/455

V = 0.081194 m3

Work done on the gas during the the process 3 to 1

W = PV = 2 x 105 x (0.081194 – 0.053535) = 5531.95 J

u = Q +W

Q = -8297.925 – 5531.95 = - 13829.875 J

Net themal energy transfer for the whole cycle = -13829.875 + 16 060.5 J

                                                                        = 2230.625 J

                                                                                    = 2231 J into the gas

Since , gas is returned to initial stage the total internal energy change is = 0

Apply first law of thermodynamics to the who cycle

U =W +Q   U = 0

W = -Q = - 2231 J    net work done (by the gas ) = 2231 J

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