Two moles of an ideal monatomic gas go through the cycle abc. For the complete cycle, 800 J of heat flows out of the gas. Process ab is at constant pressure, and process bc is at constant volume. States a and b have temperatures Ta = 200 K and Tb = 300 K.
(a) Sketch the all possible pV-diagrarns for the cycle.
(b) What is the work W for the process ca?

Respuesta :

a) Sketches of all possible pv-diagrams for the cycle are attached below

b) The work W[tex]_{ac}[/tex] for the process Ca is : 2462.8  J

Given data :

Amount of heat flowing out = 800 J

Ta = 200 K

Tb = 300 K

R = 800

B) Determine the work W for the process Ca

Wₐs = -pdv

       = - [ pVb - pVa ] ---- ( 1 )

note : pVb = nRTb ,  pVa = nRTa

Equation ( 1 ) becomes

= -nR [ Tb - Ta ]

= - 2(8.314 ) [ 300 - 200 ]

= - 1662.87

given that W[tex]_{bs}[/tex] = 0 which is isochonic

dv = 0 ( cyclic process ) = d∅ - dw

∴ 0 = 800 - ( Wₐs  +  W[tex]_{ac}[/tex] )

Therefore : W[tex]_{ac}[/tex] = 800 + 1662.8 = 2462.8  J

Hence we can conclude that the work W for the process Ca = 2462.8  J

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