Two moles of an ideal gas undergoes a closed thermodynamic cycle from state A->B
ID: 1876064 • Letter: T
Question
Two moles of an ideal gas undergoes a closed thermodynamic cycle from state A->B->C->A as shown in the diagram. From state A to state B is a constant pressure process, from state B to state C is a constant temperature process, from state C to A is a constant volume process.
(a)Find the temperatures of the gas in state A and B, respectively.
(b)Calculate the work done by the gas from state A to state B.
(c)Calculate the work done by the gas from state B to state C.
(d)How much heat is absorbed by the gas for one complete cycle.
a cosed thermodynamic cyde from state A->B-C>A as shown in the diagram. From state A to state B is a constant pressure pracess, constant volume process B Cis a constant temperature process, from state C to A is a (a) Find the temperatures of the gas in state A and B, respectively. (b) Calculate the work done by the gas from state A to state B (c) Calculate the work done by the gas from state B to state C (d) How much heat is absorbed by the gas for one complete cycle P(kPa) 2000 Y= 1.40 101 1.00 5.00W()Explanation / Answer
a)
We know that PV=nRT
So at point A,
2000000*6=2*8.314T
T=721674.2843K
And at point B,
only the volume is made 1/6
so,
T=120279.0474K.
(b)
From A to B
W=(V2-V1)P
=(-5)*2000000
=-10000000J
(c) From B to C it is a isothermal process, so
W=nRTln(V2/V1)
W=2*8.314*120279.0474*ln(6)
=3583518.939J
(d)
Now, net heat absorbed will be equal to net work done.
Q=-10000000J+3583518.939J
=-6416481.061J
that means 6416481.061J heat is absorbed.
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