A 0.2 m^3 insulated, rigid vessel is divided into two equal...

A 0.2 $\mathrm{m}^{3}$ insulated, rigid vessel is divided into two equal parts $A$ and $B$ by an insulated partition, as shown in Fig. P.12.123. The partition will support a pressure difference of $400 \mathrm{kPa}$ before breaking. Side A contains methane and side B contains carbon dioxide. Both sides are initially at $1 \mathrm{MPa}, 30^{\circ} \mathrm{C}$. A valve on side $\mathrm{B}$ is opened, and carbon dioxide flows out. The carbon dioxide that remains in $\mathrm{B}$ is assumed to undergo a reversible adiabatic expansion while there is fiov out. Eventually the partition breaks, and the valve is closed. Calculate the net entropy change for the process that begins when the valve is closed.

Key Concepts

Entropy

Entropy is a thermodynamic property that measures the disorder or randomness within a system. In processes where the system evolves from an initial to a final state, the net change in entropy helps determine the irreversibility and feasibility of the process. It is a crucial concept when evaluating thermodynamic processes, particularly in isolated systems where any increase in entropy indicates energy degradation.

Reversible Process

A reversible process is an idealized thermodynamic process that occurs so slowly and without friction or dissipative effects that the system can be returned to its original state with no net change in the entropy of the system or its surroundings. This concept is key because it provides a benchmark for the maximum possible work that can be obtained from a system and usually results in zero net entropy change.

Adiabatic Process

An adiabatic process is one in which there is no heat transfer between the system and its surroundings. In a reversible adiabatic process, the system’s entropy remains constant, whereas in an irreversible adiabatic process the entropy increases. Understanding adiabatic processes is essential for analyzing scenarios where energy changes occur solely through work interactions, as no external heat flow occurs.

Isolated System

An isolated system is one that does not exchange matter or energy with its surroundings. This concept is important in thermodynamics because in an isolated system the total energy remains constant and the behavior of entropy can be solely attributed to internal irreversible processes. Studying isolated systems simplifies the analysis of energy and entropy changes during complex processes.

Pressure Equilibration and Partition Removal

The removal or failure of a partition separating gases leads to pressure equilibration and gas mixing, which is typically an irreversible process. Understanding how pressure differences drive mixing and equilibration is key to analyzing the resulting entropy change. This concept encompasses the dynamic interactions and energy exchanges that occur when confining boundaries are altered, leading to a new equilibrium state.

Transcript

0:00 Hi, everybody.

00:01 So we need to calculate the net entropy change for the process that begins when the valve is closed.

00:09 So for this, and part b, the expansion process is reversible is idiotic.

00:15 So we have tb1 over tb2 equals kb1 over k b2.

00:28 K minus 1 over k.

00:31 Okay.

00:33 So what we can do is fill this in.

00:35 It's got 303 .15 divided by tb2 equals 1 ,000.

00:46 And your 600 in your k is been equal to 1 .289 minus 1 divided by 1 .289.

00:58 Okay.

01:00 So your tb2 is going to equal to 270 .34k.

01:08 And so now for your in your number of moles of b2, we have b2 times b2 divided by r to b2.

01:27 And so we have 600 times 0 .1.

01:32 Divided by 8 .3145 times 270 .344 equals 0 .0266k mole.

01:46 Okay.

01:49 And now we have n .a2 is going to equal to na1.

01:56 So we have 1 ,000 times 0 .1 divided by 8 .3145 times 303 .15 equals 0 .396k mole.

02:18 And now we have the energy equation as u3 minus u2 equals u3.

02:31 0, u3 minus u2b minus u2b plus u3 minus u2 of a equals 0.

02:47 Now we're going and we have nb2 times c bb times 23 minus tb2 plus na2 plus na2 times c b2 times c3 minus c b2 plus na2 times c tva times t3 minus t -a2 equals 0.

03:12 So we have 0 .0266 times 44 .01, times 0 .156 times t3 minus t -70 .34 plus 0 .0 .0 .56 times t3 minus t -70 .34 plus 0 .0 .0.

03:37 396 times 396 times 16 .0 .043 times 0 .43 times 0 .415 times t3 minus 303 .15 equals 0 .45 equals 0 .4162.

04:09 Minus 129 .274 equals zero.

04:16 And your t3 is going to be 289.

04:21 0 .722.

04:23 Let me get that 2 looking normal.

04:27 2 .2 kelvin.

04:32 Okay.

04:36 And now we have an total equals in a 2 plus.

04:47 In v2 equals 0 .0396 plus 0 .0266 equals 0 .0266 equals 0 .0662.

05:02 So now we're going to find the final pressure, which is in total times r23 of v total, the volume total.

05:16 And this is 0 .662 times 8 .3145 times 289 .72 divided by 0 .2, which is 797 .338 kilapascal...

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