An insulated vertical cylinder is fitted with a frictionless constant loaded piston of crosssect

An insulated vertical cylinder is fitted with a frictionless...

Key Concepts

Insulated System and Energy Conservation

An insulated system does not exchange heat with its surroundings, meaning that the total energy of the system is conserved. In thermodynamics, this is often referred to as an adiabatic condition. For ideal gases, whose internal energies depend only on temperature (for a given composition), if both gas mixtures start at the same temperature, energy conservation implies that the mixture’s temperature will remain unchanged even though mixing and expansion work occur.

Constant Pressure (Isobaric) Process

A constant pressure process is one in which the pressure of the system remains fixed despite changes in volume during the process. In systems that include mechanisms like a frictionless piston, the external force is adjusted so that the internal pressure is maintained. This condition influences the work done by or on the system as its volume changes, and it is a key constraint when determining final system dimensions such as the height of a gas-filled cylinder.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, provides the relationship between pressure, volume, temperature, and the amount of substance for an ideal gas. This relationship is crucial for calculating the state variables of a gas during processes like expansion, compression, or mixing. Since the internal energy of an ideal gas depends only on its temperature, the ideal gas law ties the macroscopic variables together and provides a basis for predicting how the system adjusts to satisfy constant pressure and energy conditions.

Entropy Change and Irreversible Mixing

Entropy is a measure of disorder or randomness in a system. When different gases mix, the process is usually irreversible, leading to an increase in the total entropy. Even if the overall internal energy and temperature remain constant, the mixing of gases increases the number of ways the molecules can be arranged, thus increasing the system’s entropy. This concept is central to understanding the second law of thermodynamics in the context of mixing processes.

Work and System Adjustment under Mechanical Constraints

In a system with a movable piston, changes in gas volume due to mixing result in mechanical work being done by or on the system. The work relates to changes in volume while the pressure remains constant. Even though the process is adiabatic (with no heat transfer), the work done is accommodated through volume change which can alter macroscopic dimensions like the cylinder height. This concept links the mechanical response of the system (via the piston) to the thermodynamic properties of the gases.

Transcript

00:01 Hi, everybody.

00:01 So for this one, what is the final temperature, final cylinder height in the entry change in the process? okay.

00:11 I was asking for a lot.

00:13 So the v total of the cylinder is va1 plus vb1 equals ap times the height, which is 0 .1 times 1 .0, which is 0 .1 meter cube.

00:29 Now, the volume of methane in the cylinder is a1 equals the v total minus bb1, which is 0 .1 equals 0 .0 .0 .5, which is 0 .095 meters cubed.

00:51 So the mass of methane in the cylinder is a 1 minus p .a1, ba1, divided by r.

01:00 R -a -t -1, which is going to be equal to 150 times 0 .095, divided by 0 .5183 times 300 equals 0 .0916 kilograms.

01:20 Okay.

01:21 And the neon gas, which is b1, is 500 times 0 .005, divided.

01:30 By 0 .412 times 300 equals 0 .0202 kilogram.

01:40 Now we can find the q2 in here.

01:47 But first, so we need to find the q2 and split force want to find the w2, which is going to be p pressure divided by volume, which is pressure times b2 minus b1.

02:02 And this ends up equalling to m -a -r -a times m -b, mass times the rate of b times temperature 2 minus pressure 2 times the velocity total, which is the volume total, i mean.

02:23 Okay? so now we have the q -2 is m -a -c -p -a -t -2 minus t -a -1 plus m -a -a -1.

02:33 M b, c .p.

02:38 B of t2 minus t b1 plus 1, 2.

02:43 And so here we have 0 .0916 times 2 .24 times t2 minus 300 plus 0 .02022 times 1 .0202 times 1 .0 .02 times 1 .0 .0 .2.

03:05 Times t2 minus 300 plus 0 .0916 times 0 .583 plus 0 .0202 times 0 .412 times t2 minus 15 times 0 .1 .1.

03:32 And so your q is going to equal to zero.

03:35 So now we have zero.

03:36 Equals t2 0.

03:41 T83 minus 83 .16.

03:45 And what's going to solve for a t2, and you get 293 .851 kelvin's.

03:53 Okay? and now we have to find the vf, which is going to be m -a -r -a plus m -b -r -b -t -r -b times.

04:08 T2 divided by the pressure of two.

04:11 So we have 0 .091 times 0 .5183.

04:23 I think that three correct here, plus 0 .022 times 0 .412 times 293.

04:35 0 .851 divided by pressure, so 150.

04:41 And this is going to be 0 .1093 meters cube.

04:46 Okay.

04:48 Now, we find the height of the cylinder.

04:51 So height of the cylinder is 0 .103 divided by 0 .13, and we get 1 .093 meters.

05:01 And we get 1 .093 meters.

05:03 And now we find the number of moles, okay? so number of moles of ma is 0 .0916 divided by 16 .043 equals 0 .0571 kilmoles.

05:27 And b is going to be 0 .02022 divided by 20 .183 equals 0 .0 .0 .0 .3 equals 0 .0 .0 .22.

05:39 0 .0101 kilo moles.

05:43 Okay...

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