Model air as a diatomic ideal gas with M=28.9 g / mol . A cylinder with a piston contains 1.20 k

step 1 The problem here is we have a cylinder under the spatial expansion. The expansion, pressure and

Model air as a diatomic ideal gas with M=28.9 g / mol . A...

Model air as a diatomic ideal gas with $M=28.9 \mathrm{g} / \mathrm{mol} . \mathrm{A}$ cylinder with a piston contains 1.20 $\mathrm{kg}$ of air at $25.0^{\circ} \mathrm{C}$ and 200 $\mathrm{kPa}$ . Energy is transferred by heat into the system as it is permitted to expand, with the pressure rising to 400 $\mathrm{kPa}$ . Throughout the expansion, the relationship between pressure and volume is given by $$P=C V^{1 / 2}$$ where $C$ is a constant. (a) Find the initial volume. (b) Find the final volume. (c) Find the final temperature. (d) Find the work done on the air. (e) Find the energy transferred by heat.

Model air as a diatomic ideal gas with $M=28.9 \mathrm{g} / \mathrm{mol} . \mathrm{A}$ cylinder with a piston contains 1.20 $\mathrm{kg}$ of air at $25.0^{\circ} \mathrm{C}$ and 200 $\mathrm{kPa}$ . Energy is transferred by heat into the system as it is permitted to expand, with the pressure rising to 400 $\mathrm{kPa}$ . Throughout the expansion, the relationship between pressure and volume is given by
$$P=C V^{1 / 2}$$
where $C$ is a constant. (a) Find the initial volume. (b) Find the final volume. (c) Find the final temperature. (d) Find the work done on the air. (e) Find the energy transferred by heat.

Key Concepts

First Law of Thermodynamics

The first law of thermodynamics is the principle of energy conservation applied to thermodynamic systems. It 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 (?U = Q - W). This law provides the framework for linking heat transfer, work, and changes in a system's internal energy during a process.

Polytropic Process

A polytropic process is a thermodynamic process that follows the relation PV^n = constant, where n is the polytropic index. In this context, the process is defined by a specific pressure-volume relationship (P = C V^(1/2)), which is a variation of the general polytropic form. Understanding this relationship allows one to determine how the state variables change throughout the process.

Thermodynamic Work

Thermodynamic work, in the context of gas expansion or compression, is defined as the integral of pressure with respect to volume (W = -?P dV for work done on the system). Calculating work in processes where pressure is not constant involves integrating the pressure-volume function over the change in volume, which is key to determining the energy transferred via work in such processes.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, is a fundamental equation in thermodynamics that relates the pressure, volume, and temperature of an ideal gas with the amount of gas present. It serves as the cornerstone for determining any one of these state variables when the others are known, and it is particularly useful for solving problems involving state changes in a closed system.

Diatomic Ideal Gas

Diatomic ideal gases are composed of molecules with two atoms, and they typically possess additional degrees of freedom compared to monatomic gases. This affects their internal energy and specific heats, making it necessary to consider the contributions of translational, rotational, and sometimes vibrational motions when analyzing thermal energy changes and temperature-dependent properties.

Transcript

00:01 The problem here is we have a cylinder under the spatial expansion.

00:07 The expansion, pressure and the volume will follow this equation.

00:11 So we want to find out the initial volume and the final volume temperature, final temperature, work down, and heat.

00:20 So the part a is what's the initial volume? so because this is the idea gas, so we must follow the idea gas equation.

00:31 Pv is equal to nr t so for the initial we want to find the initial value so vi initial value should be will be nr t and divided by p i and what's the end because we have 1 .2 kilogram so 1 .2 kilogram divided by 28 .9 this air times 10.

01:09 And minus 3 kilograms.

01:13 So, they want to find out the end, the more number, and the 8 .5 constant, and the temperature, 298, because the unit temperature is 25 celsius.

01:30 So, and the initial pressure is 200 par, 200 kilo par.

01:41 So finally, the initial volume will be 0 .14, 1 ,000, 0 .514 cubic meter.

01:52 Okay, this is part a.

01:56 And the part b is to find out what is the final volume.

02:03 Because the pressure and the volume has this lure.

02:08 So, from this lure, so we can follow that.

02:13 Pi squared root, vi, will be equal to pf, vf, f.

02:25 So the vf will be equal to vi, square root both sides, square root both sides, vif, pi, 0514 times pf is 300 kilo and the initial 200 kilo.

03:18 So this will be 2 .06 cubic meter.

03:28 Okay, this is part of b for the final volume of this expansion.

03:34 Now, c, what's the c? c is final, what's the final temperature? so we already know the final volume, final temperature, we should find the final temperature.

03:52 So under the ideal gas equation, pv equal to n -r -t, therefore, t -f will be equal to p -f -v -v -a, and r so this will be n will be same 1 .2 and 28 .9 this will be constant and what the pf is the 400 kilogram kilogram and what's the vf we have this is 2 .06 cubic meter so the final temperature will be 2 .3a times 103 so this is the final temperature.

04:59 D, what's the work down on the gas? so the walk down on the gas should be minus pdv.

05:11 Okay, what's the, what's the p? p is the equation will be variable.

05:27 So do the will be c, v, 1 half, d, v.

05:44 So this should be c to the integral.

06:21 So the final initial and the final volume...

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