A piston/cylinder contains helium at 110 kPa at ambient temperature 20^∘ C, and initial volume o

A piston/cylinder contains helium at 110 kPa at ambient...

A piston/cylinder contains helium at $110 \mathrm{kPa}$ at ambient temperature $20^{\circ} \mathrm{C},$ and initial volume of $20 \mathrm{L}$ as shown in Fig. $\mathrm{P} 12.119 .$ The stops are mounted to give a maximum volume of $25 \mathrm{L}$ and the nitrogen line conditions are $300 \mathrm{kPa}$, $30^{\circ} \mathrm{C} .$ The valve is now opened, which allows nitrogen to flow in and mix with the helium. The valve is closed when the pressure inside reaches $200 \mathrm{kPa},$ at which point the temperature inside is $40^{\circ} \mathrm{C} .$ Is this process consistent with the second law of themodynamics?

Key Concepts

Mixing of Gases

The mixing of different gases is a classic example of an irreversible process that inherently leads to an increase in entropy. When gases at different conditions or compositions come into contact, they spontaneously intermix, increasing the number of ways the molecules can be arranged. This process demonstrates how natural mixing phenomena comply with the second law by driving the system towards greater randomness.

Irreversibility

Irreversibility refers to processes that cannot simply be reversed to return a system to its original state without introducing changes elsewhere, such as additional changes in entropy. Many thermodynamic processes, particularly those involving spontaneous mixing or expansion, are irreversible because they create paths that depend on an overall increase in disorder.

Entropy

Entropy is a measure of the randomness or disorder within a system. In thermodynamics, processes that result in an increase in entropy are naturally favored. The mixing of substances, especially gases, typically leads to a higher combined entropy, reflecting the increase in the number of possible microstates of the system.

Second Law of Thermodynamics

This law states that the entropy of an isolated system can only increase or remain constant over time, never decrease. It governs the direction of natural processes, ensuring that energy transformations are not 100% efficient and that irreversible processes occur spontaneously in one direction, driving the overall increase in disorder in the universe.

Transcript

00:01 Hi, everybody.

00:01 So for this one, we're asked, is a process consistent with the second law of thermodynamics? so we have your q value is going to equal to n2h2 minus n1h1 plus w minus an initial, h initial, and we got n2h2, h2.

00:31 H initial minus n, initial, h initial, plus w equals n2, h2 minus n1, h1, minus n initial, plus negative b2, p2, minus p1, equals n2, h2 minus n1, h1, initial, h1, minus n initial, h1, minus b2, p2, p2, minus p1 equals n -a -2 minus h -a -1, n2 plus n -2 times h -b2 minus h -b1, and that's for your h -2 -o minus v2 times p2 minus p -1.

01:30 So now we're going to give it just a second, because we've got a find the ends so for n b is been equal to n1 which is p1 v1 r t1 equals 1110 10 .02 divided by 8 .3145 times 293 .15 equals 0 .0 .0 .0 .0.

02:01 Okay.

02:03 Okay.

02:04 And.

02:04 And now we have n2 equals an a plus nb equals p.

02:14 And we know that.

02:15 So it's going to do 200 times 0 .025 divided by 8 .3145 times 313 .15 equals 0 .092.

02:33 Kilomoles.

02:35 Okay.

02:36 And now we can do n -a, which is n2 minus nb, which is going to be 0 .00192 minus 0 .0 .09 equals 0 .0102 kilomoles.

02:55 Now back to this guy up here.

02:59 So we have q equals and it, it, equals to n -a -c -p -n2 times t -a -2 minus t -a -1, n2 for nitrogen, n -b times c -p -he for helium, tb -2 minus t -b -1, helium minus b -2, p -2 minus p -1.

03:30 Okay, and we're just going to plug in chugs.

03:32 We got 0 .001, 1 .042 times 28 .013 times 40 minus 30 plus 0 .009 times 5 .193 times 4 .003 times 40 minus 0 .009, minus 0 .02.

04:03 25 times 200 minus 110.

04:08 And this bad boy, we get 0 .2.

04:12 Let me write that a little better.

04:17 2977 plus 0 .3741 minus 2 .25.

04:24 And we get negative 1 .578 .2 kilojoules.

04:28 Okay.

04:29 And so now again, we have entropy.

04:35 Okay.

04:37 And so we have entropy.

04:42 And your entropy is equal to s -gen equals n2 -1s -1, initial, minus q of t.