A mole of ideal gas is taken from a state p1, V1 to a state p2, V2 along a path forming a straig

A mole of ideal gas is taken from a state p1, V1 to a state...

Key Concepts

Internal Energy Change for an Ideal Gas

For an ideal gas, the internal energy depends solely on temperature. The change in internal energy can thus be determined by the temperature change multiplied by the number of moles and the constant-volume heat capacity (?U = nC_V?T). This property greatly simplifies the analysis of energy changes in ideal gases.

Constant-Volume Heat Capacity

The constant-volume heat capacity (C_V) is a material-specific property that quantifies the amount of heat required to raise the temperature of a mole of a substance by one degree at constant volume. It is particularly useful in calculations involving ideal gases where the internal energy change is directly proportional to the temperature change under constant volume conditions.

First Law of Thermodynamics

The first law of thermodynamics, which states that energy is conserved, connects the change in a system's internal energy to the heat added and the work done on the system. This principle (?U = Q – W) is fundamental in understanding and calculating energy exchanges during a thermodynamic process.

Thermodynamic Work

In thermodynamics, the work done on or by a gas is given by the integral of pressure with respect to volume (?PdV) along a specific process path in a pressure–volume (P–V) diagram. This work depends on the particular path taken between the initial and final states rather than just the endpoints.

Ideal Gas Law

The ideal gas law, PV = nRT, establishes a fundamental relationship between pressure, volume, temperature, and the amount of gas. It is essential for describing the state of the gas during any thermodynamic process and provides the basis from which changes in these variables can be analyzed.

Process Path on a P–V Diagram

The specific path taken in a P–V diagram plays a crucial role in determining the work done by or on the system. For example, if the path is a straight line, there is a linear relationship between pressure and volume, allowing one to set up and evaluate the corresponding integral for work, reflecting the path-dependent nature of work in thermodynamic processes.

Transcript

00:01 In this question, we are given an ideal gas that is a mole of ideal gas that's taken from one state to another and we are asked to develop expressions for the work, the change in internal energy of the gas, and the heat transferred.

00:18 So when we start off with a pressure versus volume diagram, we're going from state p1v1 to a state p2v2 along a path forming a straight line.

00:27 So it doesn't really matter which direction this goes because we are going to develop a general relationship.

00:41 And we're going to dash across this way just to make it a little bit clearer.

00:46 So we're going to let this point be v1, p1, right? that's supposed to be 1, sorry.

01:03 And then we will let this point up here be v2, p2.

01:14 So i chose this arbitrarily.

01:17 You could have it be a line that slopes up and to the right as we connect the two to indicate the direction of the process.

01:31 So we need to develop a relationship for the work done.

01:37 Well, this looks like a trapezoid and we're going to remember that work done equals the area on a pv diagram.

01:56 And since this looks like a trapezoid, we can do the average of the bases multiplied by the height of the trapezoid.

02:08 Well the average of the bases is 1 half and then our two bases are going to be p1 and p2, our vertical sections.

02:19 And then our height of the trapezoid is the horizontal section where we end at v1 and start at v2.

02:28 So we'll do v1 minus v2.

02:31 And now here's the thing to consider is that this is a compression and so delta v is actually going to be negative because the gas ends up taking up less volume.

02:46 But this means something external did the work to compress the gas.

02:50 So we need this to end up with a positive value, right? because we say work done on the gas is positive because it increases the gas's energy.

03:05 Work done by the gas is going to be negative because that will decrease the gas's energy.

03:11 So that means that for our work expression we just simply need a negative sign out here so that we have a negative to multiply by the negative delta v.

03:20 And next let's find an expression for delta u.

03:30 So there's mention made of the specific or the constant volume heat capacity cv and the relationship between u and cv is cv times temperature which gives us nkbt over gamma minus 1.

03:57 So we can infer from this that delta u is going to be nkb delta t over gamma minus 1.

04:10 So for this specific gas, i'm not going to substitute n yet, but we'll have nkbt2 minus t1 and then divided by 0 .4 because gamma was 1 .4 which minus 1 is 0 .4.

04:31 And now from here we are able to use the first law of thermodynamics that delta u equals the heat transfer plus the work done to say that q will be delta u minus w and i'm not going to copy these other terms back down.

04:59 I'm just going to leave it like this for our purposes.

05:02 And so now we're given some initial and final states for the gas.

05:08 We are given that we actually don't have p1.

05:18 We're going to need to find it, but we're given that v1 equals 21 times 10 to the minus third cubic meters and that t1 is 290 kelvin.

05:36 P2 is also something we're going to need to find, but we are given that v2 is 22 times 10 to the minus third cubic meters.

05:45 So this is an expansion and then t2 is 330 kelvin.

05:53 So the process that this is going to follow isn't going to look exactly like what we have pictured here in the graph.

06:01 It's actually going to go the other way with an expansion.

06:03 So we should expect our value for work to be negative...