A thermally insulated chamber contains some hot gas and a lump of metal. Initially the gas and t

A thermally insulated chamber contains some hot gas and a...

Key Concepts

First Law of Thermodynamics

This fundamental principle states that the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system. It allows us to account for energy exchanges in processes, whether they involve heat transfer or work, and is essential for analyzing any thermodynamic process as it highlights the conservation of energy in isolated or interacting systems.

State Functions

State functions are properties of a system that depend only on the current state, not on how that state was reached. For example, internal energy and temperature are state functions, meaning that regardless of the process path, the change in these properties depends solely on the initial and final states. This concept is crucial when applying energy conservation laws, as it often simplifies the analysis by allowing comparisons of states without detailed knowledge of the path taken.

Quasistatic Process

A quasistatic process is one that occurs infinitely slowly so that the system remains in near-equilibrium at all times. This assumption allows for the accurate application of thermodynamic equations since the system can be treated as being in equilibrium at every infinitesimal step. Quasistatic processes are idealizations that facilitate the analysis of thermodynamic reactions in systems where pressures, volumes, and temperatures adjust continuously.

Isothermal Process

An isothermal process is one during which the temperature of the system remains constant. In such a process, any heat added to the system is exactly balanced by the work done by the system, such that the internal energy remains unchanged for an ideal gas. This concept is essential when analyzing processes where thermal contact ensures temperature uniformity over the course of expansion or compression.

Adiabatic Process

In an adiabatic process, no heat is exchanged with the surroundings, so any changes in the system's internal energy are solely due to work done on or by the system. This process often results in temperature changes within the system itself. Adiabatic conditions are important in thermodynamics for understanding rapid processes or those where insulation prevents heat transfer, contrasting them with isothermal processes where temperature is held constant.

Pressure-Volume Diagram

A pressure-volume (P-V) diagram graphically represents the relationship between the pressure and volume of a system during a thermodynamic process. It is an essential tool that helps visualize the work done by or on the system, as the area under the process curve corresponds to the work involved. Such diagrams are key for comparing different thermodynamic paths and understanding how changes in pressure and volume affect the energy state of the system.

Transcript

00:01 In this question we have a closed cylinder with a movable piston dividing two monotonic ideal gases into separate compartments and we are going to determine some things about those two gases, then add some heat to one of them and determine the final volume and temperature for both of them.

00:28 So how many moles of gas are in each compartment? well let's first of all note that they have equal volumes and that's our 5 .0 times 10 to the minus second cubic meters.

00:52 We're going to need to put our temperature of zero degrees celsius into kelvin by adding 273.

01:11 So this will just be a temperature of 273 kelvin and it may also be helpful since we're starting off at a pressure of one atmosphere to express that in terms of pascals and we can remember that one atmosphere is 1 .013 times 10 to the fifth pascals.

01:49 So to determine the number of moles in each of each gas, well they're at the same pressure, they're at the same volume, they're at the same temperature, so we only have to calculate this once.

02:07 We are just going to use the ideal gas law pv equals nrt.

02:12 So we can solve this for r, sorry for n by dividing by r and t.

02:39 So we can substitute now our pressure 1 .013 times 10 to the fifth, our volume 5 .0 times 10 to the minus second, and our gas constant has a value of 8 .314 and of course now multiplied by 273 kelvin and i get 2 .225 moles.

03:44 Now in part b we're going to add some heat slowly to a so it expands and then b is compressed and this is going to happen until the pressure of both gases is three atmospheres.

03:56 And we're given this key clue here that the compression of b is adiabatic and we'll use this to determine the final volume of both gases.

04:10 So first step here for part b is let's express our new temperature.

04:18 I'm going to call it p2 because of course we had p1 which i'll come up and label p1.

04:28 So p2 is three atmospheres which converted into pascals in case we need it in pascals.

04:41 When you use the full number of decimal places for that conversion factor there are a few more decimals after the 0 .013.

04:54 You end up with 3 .04 times 10 to the fifth pascals.

05:10 Oh i should also put pascals up here.

05:12 There we go we'll fix that.

05:13 Okay so the equation for handling the adiabatic expansion or compression of a gas is p1 times v1 raised to the factor of gamma equals p2 times v2 raised to the factor of gamma.

05:46 And this factor of gamma is the molar specific heat capacity at constant pressure cp divided by the molar specific heat capacity at constant volume cv.

06:05 Oh my goodness we weren't given those things but this value of gamma is equal to five thirds for monatomic ideal gases which is what we are working with.

06:23 And this was something i had to go do a little bit of research on to find.

06:26 This was not covered in at least not in this current chapter of the textbook that this question came from so i had to go consult some other sources in order to find this out.

06:41 So now that we have now that we have our power of gamma we can go ahead and calculate the new volume for b and then subtract that from the total volume of the container to get the new volume for a.

07:06 So i'm going to modify my equation just a teeny tiny bit.

07:10 I'm going to write it instead as p1 b times v1 for b raised to the power of gamma equals p2 for b times v2 for b raised to the power of gamma.

07:33 I'm going to go ahead and solve this symbolically before i substitute.

07:51 So i'll divide both sides by p2 b and now you know remember that we're going to do an inverse operation here.

08:08 So since v was raised to the power of gamma we are going to need to take the one over gamma root.

08:17 So since gamma was five -thirds we're going to find the three -fifths root of p1 b times v1 b raised to the gamma power divided by p2 b.

08:46 Sorry i should cut that off.

08:50 So let's keep going with this substitution.

09:03 So again p1 b.

09:04 Well here's the thing.

09:06 Since we are dividing the original pressure by the end pressure it doesn't actually matter which units we use because it will be the same units.

09:17 I mean the units will cancel out.

09:20 So i'm going to be a little bit lazy.

09:22 I'm just going to use one atmosphere times that original volume of 5 .0 times 10 to the minus second cubic meters.

09:38 That will be raised to the five -thirds power then divided by three.

09:49 Because again guys we you know we have that factor to give our atmospheres to pascals and then we multiply by three so that factor ends up canceling out as well in the algebra.

10:04 So this just makes it nice and short.

10:07 All right i get a value whether i express it like this or the ratio of p1 to p2 to the three -fifths power multiplied by v1.

10:21 Doesn't matter how i plug it in.

10:22 I'm going to get 2 .586 times 10 to the minus second cubic meters...

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