Consider the adiabatic free expansion of a gas since there is no external Pressure hence Work done on the system is 0 and since the walls are insulated (hence adiabatic) the heat absorbed is 0. However since this is a irreversible process then entropy change > 0 hence dQ > 0 . However there is no heat absorption. What am I missing ?
Thermodynamics – Analyzing Entropy Change in the Free Expansion of a Gas
adiabaticentropythermodynamics
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The ideal gas equation only applies to an ideal gas in a thermodynamic equilibrium state, and not to one that is rapidly deforming. The 2nd equation you wrote applies only to two closely neighboring thermodynamic equilibrium states of a system, and describes the relationship between the differential changes in U, S, and V between these adjacent states. Neither equation describes the intermediate changes along an irreversible path between two widely separated thermodynamic equilibrium states. So how do you determine the change in entropy between these two end states that are connected in practice by an irreversible path?
Step 1: This is the most important step. Completely forget about the actual irreversible path between the two end states. Focus only on the initial and final end states.
Step 2: Devise a reversible path between the two end states. The reversible path you devise does not have to bear any resemblance whatsoever to the actual irreversible path, as long as it starts and ends at the same two end states. A reversible path consists of a continuous sequence of thermodynamic equilibrium states. So, along the path you choose, the system can be no more than slightly removed from thermodynamic equilibrium at every point along the path. There are an infinite number of reversible paths than can take you from state A to state B. So choose one that is convenient for performing step 3.
Step 3: Calculate the integral of dQ/T for the reversible path that you devised in step 2.
When they say that the change in entropy is equal to the integral of $dQ_{rev}/T$, what they mean is that you have to calculate the integral for a reversible path.
You can alternatively use your equation 2 to determine the change in entropy since it automatically implies a reversible path.
You are assuming, wrongly, I think, that (1) and (3) can be joined by an adiabatic curve. For an ideal gas, two points which can be gone between by a reversible adiabatic process are related by $$p_a\ V_a^\gamma=p_b\ V_b^\gamma$$ in which $\gamma=\frac{c_p}{c_V}$, so in this case$$\frac{p_3}{P_{1,2}}=\left(\frac{V_{1}}{V_{2,3}}\right)^\gamma$$
But we know by applying the ideal gas equation of state, $pV=nRT$, that$$\frac{p_3}{P_{1,2}}=\frac{300}{700}\ \ \ \text{and}\ \ \ \frac{V_1}{V_{2,3}}=\frac{500}{700}$$ We can show that, for any gas, $1.00<\gamma<1.67$
But $\left(\frac{500}{700}\right)^1=\frac{500}{700}\ \ \ \ \ $ and $\ \ \ \ \left(\frac{500}{700}\right)^{1.67}=0.57=\frac{400}{700}$
So it's never going to work!
Best Answer
Entropy can be generated without there being heat transfer, i.e., when $Q=0$. That's the case for a free expansion into a vacuum. The classic example given is an ideal gas located in one side of a rigid insulated vessel with a vacuum in the other side separated by a rigid partition. An opening is created in the partition allowing the gas to expand into the evacuated half of the vessel. $W=0$, $Q=0$, $\Delta T=0$ (for an ideal gas) and therefore $\Delta U=0$. Although no heat transfer has occurred, the process is obviously irreversible (you would not expect the gas to be able to spontaneously return to its original location) and entropy increases.
You can calculate the entropy increase by assuming any convenient reversible process that can bring the system back to its original state (original entropy). The obvious choice is to remove the insulation and insert a movable piston. Then conduct a reversible isothermal compression until the gas is returned to its original volume leaving a vacuum in the other half. All properties are then returned to their original state. The change in entropy for the isothermal compression is then, where $Q$ is the heat transferred to the surroundings by the isothermal compression,
$$\Delta S=-\frac{Q}{T}$$
Since the system is returned to its original state, the overall change in entropy is zero, meaning the original change in entropy due to the irreversible expansion has to be
$$\Delta S=+\frac{Q}{T}$$
Hope this helps.