A 0.05-m^3 well-insulated rigid tank contains oxygen at 175 K and 6 MPa. A paddle wheel placed i

A 0.05-m^3 well-insulated rigid tank contains oxygen at 175...

Key Concepts

Ideal Gas Law

The ideal gas law (PV = nRT) relates the pressure, volume, temperature, and amount of gas in a closed system. It is a fundamental equation used to determine the state of a gas under various conditions. In problems involving temperature changes in a rigid container, the ideal gas law is used to calculate the final pressure after the temperature has been altered.

Paddle-Wheel Work

Paddle-wheel work is a form of work interaction where mechanical energy is supplied to the fluid via a rotating paddle wheel. This is a type of shaft work that can increase the internal energy of the system without affecting the volume, distinguishing it from pressure–volume work. It is particularly important in processes where mechanical agitation leads to changes in the thermodynamic state of the system.

Internal Energy Change in an Ideal Gas

For an ideal gas, the change in internal energy is primarily a function of temperature and can be calculated using the specific heat at constant volume (cv) multiplied by the temperature change. This concept is crucial for processes in which the internal energy change is used to determine other work or energy transfer quantities, especially in systems with no changes in kinetic or potential energy.

Rigid Tank Process

A rigid tank indicates a closed system where the volume remains constant throughout the process. In such a system, no boundary work is done because the volume does not change, meaning that any energy change in the system is not due to expansion or compression work. This concept is essential in energy balance analyses where work interactions must be carefully distinguished from volume-related work.

First Law of Thermodynamics for Closed Systems

For a closed system, the first law of thermodynamics states that the change in internal energy is equal to the net heat transfer into the system plus the work done on the system (?U = Q + W). In an insulated, rigid system, this principle helps in accounting for energy changes solely due to work interactions, since there is no heat exchange with the surroundings.

Transcript

00:01 For this problem on the topic of thermodynamic property relations, we are told that a rigid tank that is well insulated contains oxygen at a given state.

00:10 A paddle wheel is then placed in the tank and turn on, which increases the temperature of the oxygen to a higher temperature which is given.

00:19 We are to use the generalized charts to determine the final pressure in the tank as well as the work done by the paddle wheel in this process.

00:29 So firstly, we will use the critical properties, as well as the compressibility factor and the entire be departure factors.

00:39 And we'll first calculate the reduced temperature tr1.

00:45 So tr1 is equal to the specific temperature t1 over the critical temperature tc.

00:52 So this is 175 divided by the critical temperature for oxygen, which is 154 .6.

01:02 This gives us the reduced temperature of 1 .13.

01:07 The reduced pressure in the initial state, pr1, is equal to the specific initial pressure p1 over the critical pressure for oxygen pc.

01:17 And this is 6 divided by 5 .043.

01:23 This gives us a reduced pressure of 1 .19.

01:28 Now, from these values, if we go to our generalized charts, we get a compressibility factor.

01:34 Factor z1 of 0 .682 as well as in dalpi departure zh1 in the initial state of 1 .33.

01:48 Now we also know that p times v is equal to zrt, which means we can rearrange and solve for the specific volume, v1.

02:04 So v1 is equal to zrt over p.

02:11 So that's z1, which is 0 .682 times r, the gas constant for oxygen, which is 0 .2598, in si units, multiplied by the temperature t1, 175 kelvin.

02:39 This is divided by the pressure p1, which is 6 ,000 kilopascals.

02:45 And this gives us a specific volume v1 of 0 .00516 cubic meters per kg.

02:55 So that's the initial specific volume of the oxygen.

03:00 Therefore the mass of the oxygen can be calculated as follows.

03:03 The mass is the total volume, capital v, over the specific volume little v1.

03:09 That's 0 .05 cubic meters divided by the specific volume 0 .00516 cubic meters per kg.

03:24 We can see we're left with units of mass and we get the mass of oxygen to be 9 .68 kg.

03:34 Now the specific volume of oxygen remains constant throughout this process...

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