When we remove the stoppers, the gas expands and the piston shoots up and eventually reaches a new final position in which the internal and external pressures are the same.
Apparently we can write
$$\delta q=0\tag{1}$$
$$\delta w=-P_2dV\tag{2}$$
$$dU=C_VdT\tag{3}$$
$$dU=-P_2dV\tag{4}$$...
It is clear that the process is isothermal else it is not possible to find heat absorbed.
$$W=-P_{ext}(\Delta V)$$
However ##P_{ext}## is not given. How do I proceed?
I tried taking ##W=-(P_2V_2-P_1V_1+P_3V_3-P_2V_2)=\Delta(PV)## but it is wrong for obvious reasons.
I came across the following statement from the book Physics for Engineering and Science (Schaum's Outline Series).
I cannot seem to find a satisfactory answer to the questions.
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Hi all. I am referencing the example given in Halliday and Resnick, Chapter 20, Section 1, Subsection "Change in Entropy". The above picture is graph of the free expansion of a gas into a volume that is double its original volume. I n a free expansion there is no heat transfer, the pressure...
Attempt at A Solution
Problem 1
Reversible Process - A cylinder of ideal gas at pressure P is in mechanical equilibrium with a piston of area A driven by a spring of force F = PA and thermal equilibrium with a reservoir of temperature T. The piston is moved a small distance dx toward the...
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From a video lecture, it is mentioned that "dU≠dW in Joule's free expansion if the process is irreversible and adiabatic"
Mentioned in around 36:00-38:00 in the video:
What I would like to ask is why in this irreversible adiabatic process, dU≠dW? Is it because the W here doesn't include other...
The equation for entropy S=delta(Q)/T is derived from reversible processes such as Carnot cycle. The delta(Q) in the equation is the reversible heat added or taken out from the system. So, why is this equation valid in the case of processes like cooling of a body which is irreversible?
I know that gibbs free energy for say a body will be equal to = Gibbs free standard energy at 1M and Ph7(-rtlnkeq) (Where k is the concentration of product/ concentration of reactant at equilibrium)+rtlnk.
How can we use the standard gibbs free for irreversible spontaneous processes? Is it...
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he explain about Reversible and Irreversible processes by using "The number of accessible states in equilibrium(Ω)".
in his point of view, each accessible states have equal probability.
comparing to weather...
I was following along in my Thermodynamic textbook and began playing with some definitions. In the following formulation, I somehow managed to prove (obviously incorrectly) that dq = TdS for even irreversible processes. I was hoping someone could point out where in the proof I'm going wrong...
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I'm working on a problem of the thermal stability of a protein. Conventionlly, people compare protein thermal stability in terms of the Gibbs free energy difference between the native and unfolded state. So if it reversibly falls apart, then for N <==> U, DG(N-U) is accessed from the...