Find ΔE, ΔH, q, and w for the change in state of 1.0 mol H2 O(l) at 80^∘ C to H2 O(g) at 110^∘ C

Find ΔE, ΔH, q, and w for the change in state of 1.0 mol H2...

Key Concepts

Enthalpy

Enthalpy is a thermodynamic state function defined as the sum of a system's internal energy and the product of its pressure and volume (H = E + PV). Under constant pressure, the change in enthalpy (?H) corresponds directly to the heat absorbed or released by the system.

Work

Work in thermodynamics is the energy transferred when a system changes its volume against an external pressure. Under constant pressure conditions, it is often calculated as w = -P?V, reflecting the energy required for expansion or compression of the system.

Latent Heat

Latent heat is the energy absorbed or released during a phase change, such as from liquid to gas, at constant temperature and pressure. It represents the energy needed to overcome intermolecular forces during the phase transition without a change in temperature.

First Law of Thermodynamics

The first law states that energy is conserved; the change in internal energy of a system is the sum of the heat exchanged and the work done (?E = q + w). This principle is fundamental in analyzing energy changes during physical transformations.

Internal Energy

Internal energy refers to the total microscopic energy contained within a system, including both kinetic and potential energies of its particles. Changes in internal energy provide insight into the net energy transfer through heat and work during a process.

Heat Capacity

Heat capacity is the amount of heat required to raise the temperature of a substance by one degree. It allows for the calculation of the total heat exchanged during temperature changes, often through integration over the temperature interval involved.

Transcript

00:01 So for this problem, we're going to be heating up some water, vaporizing it, and then heating up the vapor a little bit.

00:06 And we want to know what the thermodynamic properties are.

00:09 One way we can do this is we can keep track of everything by using a chart here, where the x -axis is going to be the heat flow of the system, and the y -axis will be the temperature of the system.

00:20 In this case, the water or the water vapor.

00:23 So i'm going to start out with a certain temperature, and we'll say that this is, we're starting out at 80 celsius.

00:31 Then we heat it up to the boiling point, which is at 100 celsius.

00:39 And then we'll do a phase change.

00:42 Phase change has constant temperature, but i'm adding heat to the system.

00:46 And then we'll heat it up a little bit more until we get to our final temperature of 110 degrees celsius.

00:54 So let's go back here and check the phases.

00:56 We start out in the liquid phase.

00:58 Here, we're still in the liquid phase.

01:00 And then the evaporation happens where we transit to the gas phase, and then we stay in the gas phase.

01:08 So i'm going to refer to these as steps one, two, and three.

01:15 So let's go ahead and figure out what the heat is for step one.

01:20 So heat of step one, q is equal to number of moles, smaller heat, change in temperature, and we're told that we have one mole.

01:35 We're told the molar heat for water.

01:39 Remember this is in the liquid phase still.

01:42 So it's 75 .3 joules per mole kelvin.

01:50 And then our change in temperature is 100 celsius minus 80 celsius.

01:58 Now watch after your units here.

02:00 So moles, moles.

02:02 And really the kelvin and the celsius do cancel.

02:06 It's maybe a bit hard to see, but we've shown before that if you have a change in temperature, change in kelvin is equal to change in degrees celsius.

02:16 So those turn out to be the exact same, but only if you have delta t.

02:19 If you have t all by itself, it doesn't work.

02:22 So multiply all this together.

02:24 We get 1 ,506 joules for step one.

02:31 For step two, we can, use the equation heat is equal to number of moles times the enthalpy of vaporization because that is the process that is happening again this is sort of from chapter 11 but we can sort of work with it a little bit in chapter 6 so the numbers again we have one mole and we're given the entropy of vaporization is 40 .7 times 10 to the 3rd joules per mole and we get for an answer 40 .7 times 10 to the third joules.

03:14 Okay, step three, we're heating up a gas.

03:18 So we're going to use q equals nc delta t once more, but our specific heat will be different, or a molar heat will be different, because now we're dealing with a gas instead of a liquid.

03:29 So this is equal to one wall times the molar heat, which we're given to be 25 .0 joules per mole kelvin.

03:44 In the change in temperature this time, we end up at 110 celsius minus the initial temperature, 100 celsius.

03:52 And so we get for an answer, 250 joules.

03:59 So adding everything up together, steps 1, 2, and 3.

04:02 The q total here is going to be equal to 4 .25 times 10 to the 4th joules.

04:13 One other thing to note here is we're working under constant pressure.

04:17 If we just have some water evaporating, the atmosphere doesn't actually change its pressure because of this.

04:23 And so we're going to say that the entropy of favor, sorry, the change in enthalpy here at constant pressure is just equal to the heat.

04:32 And so it's four, it's also 4 .25 times send to the fourth joules because we're working at constant pressure.

04:43 Okay, now let's do the work.

04:45 So the work, for one, remember, work is going to be dictated by a change in volume...

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