A certain ideal gas has molar heat capacity at constant...

A certain ideal gas has molar heat capacity at constant volume $C_{V}$ . A sample of this gas initially occupies a volume $V_{0}$ at pressure $p_{0}$ and absolute temperature $T_{0-}$ The gas expands isobarically to a volume 2$V_{0}$ and then expands further adiabatically to a final volume 4$V_{0-}$ (a) Draw a $p V$ -diagram for this sequence of processes. (b) Compute the total work done by the gas for this sequence of processes.(c) Find the final temperature of the gas. (d) Find the absolute value $|Q|$ of the total heat flow into or out of the gas for this sequence of processes, and state the direction of heat flow.

Key Concepts

First Law of Thermodynamics

This foundational principle of energy conservation states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In analyzing thermodynamic processes, this law helps determine the distribution of energy between heat transfer and work, providing insights into how energy is conserved during various types of process changes.

Work Done by a Gas

Work in thermodynamic processes is defined as the energy transferred by the system when it expands or contracts, calculated by the integral of pressure over a change in volume. For isobaric processes, the work is simply the product of pressure and the change in volume, while for adiabatic or other variable pressure processes, integration of p dV is required using the relevant state equations or process-specific relations. This concept is key to linking mechanical work with energy changes in the system.

Heat Capacities

Heat capacity is a measure of how much heat energy is needed to raise the temperature of a substance by a given amount. The molar heat capacity at constant volume (CV) and at constant pressure (CP) differ due to the work done by the gas in expansion processes. Their relationship, given by CP = CV + R for an ideal gas, is essential for understanding how energy is partitioned between increasing the internal energy and doing work during processes that involve changes in temperature.

Ideal Gas Law

This concept describes the relationship between pressure, volume, and temperature of an ideal gas through the equation pV = nRT. It is fundamental in connecting the state variables of a gas and is used to determine how a gas behaves under various processes, such as isobaric or adiabatic changes. By applying this law, one can relate changes in volume and temperature while keeping the pressure constant, or vice versa.

Thermodynamic Processes

In thermodynamics, different processes define how a system exchanges energy with its surroundings. Key processes include isobaric (constant pressure) and adiabatic (no heat exchange) processes. Each process has its distinct characteristics: in an isobaric process, the work done is directly related to the change in volume, while in an adiabatic process, the relationship between pressure and volume is governed by a specific adiabatic exponent. Understanding these processes is crucial for analyzing energy transformations in gases.

Transcript

00:01 So in this problem we have a certain guess that is going through two distinct processes that are connected in time, but the first process is going to happen under constant pressure and this is going to be an expansion from v0 to 2v0.

00:18 This is, for example, here, this constant pressure, starting constant pressure, and we have two points.

00:30 Point here and point here and here and here will be a straight line that is parallel to the volume axis now from this point over 2b0 the gas will enter an adabatic expansion meaning that the pressure will drop and the volume will increase but adiabatic so from here to to here, for example, the pressure will not drop much, but it will drop.

01:10 It will be a curve, a diabetic curve.

01:15 It goes something like this.

01:18 So two processes.

01:20 First process here and the second process here.

01:30 Can also say this is the point one, the point two, and the point three.

01:36 And this is for the part a of this problem to sketch the pressure volume by aground.

01:43 Now for the part b of the problem, we need to find a total work done.

01:49 The total work done is equal to the addition of the first process, the work corresponding to the first process, which is going to be an isobran process, so w1 plus and next it is an adiabatic process.

02:07 And so the work will correspond to the work of an adiabatic process, wt.

02:14 Here we should substanti the corresponding formulas.

02:19 The work done in an isobric process in this case will be equal to p -0 times the final volume 2v0 minus v0 and for the adabatic process from the textbook we know that the formula for the work done is going to be c v over r times and here is initial pressure times initial volume is 0 2 v0 minus now we have p3 this is pressure corresponding to the third state the third point of the pv diagram and for v0 which is the final volume so here we can do some further substitutions knowing that that for an adiabatic process, the pressure p3 over p initial is going to be equal to the relationship between the initial volume and the final volume in this ratio over gamma.

03:34 So this implies that p3 equals p0 times and we can obviously see that v zeros will cancel out but this is for the later part 2v0 over 4v0 gamma.

03:54 This can be substituted here.

04:02 And when we do this, we obtain, since we are asked to find the overall expression for this case, wt equals v0, 2v0 minus v0 plus, and then we say, over r and then we have b0, 2v0, and then we have to substitute actually minus p0 and then 2v0 over 4v0 on gamma times 4 v0.

04:56 When we cancel out all of the v zeros and when we, when i mean, when i say that, i mean, all of them in the whole expression here, so from here to here.

05:11 And then we arrange it, so we place common multiplicators outside of the brackets.

05:22 We obtained that the final expression will be v0 v0, 1 plus cv over r, and then 2 minus 2, 2 minus gamma.

05:43 So this 2 minus gamma is obtained when we arrange this part here and just the 2 is from this 2 right here.

05:59 And then the p0 goes in front.

06:02 And then this p0 goes in front as well.

06:05 And of course here we have only 1 with 0...

Please give Ace some feedback