Find expressions for and for an ideal gas of given γ. |...

Name: Problem 8, Chapter 7: Thermodynamics: A complete undergraduate course
Uploaded: 2021-12-10T07:39:22-08:00
Duration: 8 min 58 s
Channel: Christina Bauer
Description: Find expressions for and for an ideal gas of given γ.

Key Concepts

Adiabatic Index (?)

The adiabatic index, commonly represented by ?, is the ratio of specific heats at constant pressure and constant volume (? = c_P/c_V). It characterizes the specific heat properties of a gas and plays a crucial role in determining the relationship between pressure and volume during adiabatic processes. For an ideal gas, ? is used to relate the adiabatic compressibility to the isothermal compressibility.

Ideal Gas

An ideal gas is a theoretical model in which the particles of the gas do not interact except through elastic collisions and occupy negligible volume. Its behavior is described by the ideal gas law, PV = nRT, which relates the pressure, volume, and temperature of the gas. This foundational model simplifies calculations and underpins many thermodynamic derivations and concepts.

Isothermal Compressibility

The isothermal compressibility, denoted as ?_T, measures the relative change in volume of a substance in response to a change in pressure while keeping the temperature constant. It is mathematically defined as ?_T = - (1/V)(?V/?p)_T and is a key parameter in describing how compressible a material is under isothermal conditions.

Adiabatic Compressibility

The adiabatic compressibility, denoted as ?_S, measures the relative change in volume of a substance in response to a change in pressure when no heat is exchanged with the surroundings (i.e., under adiabatic conditions). It is defined similarly to isothermal compressibility but with the derivative taken at constant entropy, ?_S = - (1/V)(?V/?p)_S. This parameter is important in understanding the behavior of gases during rapid processes where heat exchange is negligible.

Transcript

00:01 In this problem, we're asked to solve for alpha and kappa.

00:05 These are two separate coefficients.

00:07 Alpha is a cubic expansion coefficient.

00:12 Meanwhile, kappa is the isothermal compressibility.

00:15 And these are their formulas.

00:17 They have a lot of similarity in that they both involve a partial derivative.

00:23 Remember, the squiggly ds mean partial.

00:28 We're still taking a derivative in a normal way.

00:32 But it implies that there's more than that there's more them one variable that's changing.

00:38 So to solve these, we're keeping one variable constant changing the other.

00:44 So for alpha, our t is changing, our p is constant.

00:50 For kappa, our p is changing, and our t is constant.

00:55 So we're partially differentiating.

00:58 In both cases, they're multiplied by a negative 1 over v for kappa or a 1 over v for alpha.

01:06 Now, we're asked to solve this, we're asked to solve for alpha and kappa for a guess that follows this equation.

01:21 This is a simpler version of the vanderval's equation.

01:28 B is a van derval's coefficient.

01:37 The full van dervals equation involves a correction for b and for a.

01:42 So that's why i refer to this as a partial.

01:45 It just has b.

01:47 So in order to solve for alpha and kappa, we need to take the durific.

01:51 So first we need to organize this equation so that it is an expression in terms of v.

01:59 Both derivatives involve derivative of v with respect to different variables.

02:06 So our first goal is to find out what dvdtp is.

02:16 So i'm going to take this expression and rearrange it for so we're taking our gas equation.

02:24 V minus b equals r t over p so v is equal to r t over p plus v so this is the form we want to work with now let's evaluate dv over d tp looking at this formula if p is a constant we're essentially left with r over p is our constants and t is our variable so just like the derivative of a, x is a, the derivative of rt over p, where p and r are constants, is simply r over p.

03:20 We are not including the b factor because this does not involve temperature.

03:30 And when we take the derivative of a constant, we get zero...

Please give Ace some feedback