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# So in isobaric the volume is going to go

like so ------->----- then isothermal is a curved line, and then the volume will decrease and go straight down from isochoric. I'm not sure what to do from there or if that is even right. Any help would be appreciated thanks.

PC.L2.4-ﬂl 3.0 moles of an ideal gas are subiected to the following processes. First the volume is
tripled in an isobaric process. Then it undergoes an isothermal process to a pressure of
9.0 kPa. The volume is then halved in another isobaric process. Finally, it returns to
the original state in an isochoric process.
(a) Draw a PV diagram of the cycle. Label each state (vertex) with a letter (A, B, . . .) and each
transition with a number and arrow showing the direction of the process.
([3) If the isothermal process occurs at a temperature of 900 K5 what is the pressure5 tempera—
ture, and volume for each state.
(c) Find the work done on the gas, the heat in and out of the gas, and the change in thermal energy for each of the processes.

I have tried to solve step-by-step. Thanks... View the full answer

initially assumed
Finally got this one
(9) ( KPa) PTT
A
A
D
9kpa
3V V
2 V
VI
2
D
C
and v1 =V
V / 2 2 V
P
2
A.
B
- GH
A B - C ,isothermal Process
3 V
PV = constant
: .
PB VB = PC VC
PR ( 3 v ) = ( 9...

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