Internal Energy, Heat, and Work Thermodynamics, Pressure & Volume, Chemistry Problems

Summary

This educational video script explores chemistry problems involving internal energy, heat, and work. It explains the first law of thermodynamics, using the formula ΔU = q + w, where ΔU is the change in internal energy, q is heat absorbed or released, and w is work done on or by the system. The script clarifies that q is positive for heat absorption and negative for release, while w is positive when work is done on the system and negative when work is done by the system. It provides examples to calculate internal energy changes in various scenarios, including heat absorption, work done, and gas expansion or compression against external pressure. The script also covers the conversion of work units from liters times atm to joules.

Takeaways

• 🔍 The video focuses on chemistry problems involving internal energy, heat, and work, emphasizing the application of the first law of thermodynamics.
• 🌡️ The change in internal energy (ΔU) is calculated using the formula ΔU = q + w, where q is the heat energy and w is the work done on or by the system.
• ♨️ Heat energy (q) is positive when absorbed by the system (endothermic process) and negative when released by the system (exothermic process).
• 🔨 Work (w) is positive when done on the system, increasing its internal energy, and negative when done by the system, decreasing its internal energy.
• 📈 The video explains the sign conventions for q and w in relation to the system and surroundings, crucial for correctly calculating ΔU.
• 🧪 Two examples are provided to illustrate the calculation of ΔU, one where the system absorbs heat and work is done on it, and another where the system releases heat and does work on the surroundings.
• 🌐 The video uses visual aids like diagrams to help understand the transfer of energy between the system and surroundings.
• 📉 The formula for work done by a gas during expansion or compression is w = -pΔV, where p is the constant external pressure, and ΔV is the change in volume.
• 🔄 The concept of energy storage in gases through pressure is discussed, highlighting that compressing a gas increases its internal energy, while expansion decreases it.
• 🔗 The video provides a conversion factor between liters-atm and joules to facilitate the calculation of work in terms of energy.
• 🔋 A final example calculates the internal energy change of a gas that absorbs heat and expands against a constant external pressure, demonstrating the application of the concepts discussed.

Q & A

What is the change in internal energy (ΔU) of a system when 300 joules of heat energy is absorbed and 400 joules of work is done on the system?

The change in internal energy (ΔU) can be calculated using the formula ΔU = q + w. Since 300 joules of heat is absorbed (q = +300 J) and 400 joules of work is done on the system (w = +400 J), the change in internal energy is ΔU = 300 J + 400 J = 700 J.

How does the sign of q (heat) affect the internal energy of a system?

In the context of the first law of thermodynamics, q is positive when heat is absorbed by the system (endothermic process), leading to an increase in internal energy. Conversely, q is negative when heat is released by the system (exothermic process), resulting in a decrease in internal energy.

What is the significance of a positive w (work) in thermodynamics?

A positive w indicates that work is done on the system, which means the system's internal energy increases. This typically occurs when the system is being compressed or when an external force is applied to it.

If a system releases 700 joules of heat and does 300 joules of work, what is the change in its internal energy?

The change in internal energy (ΔU) for this scenario would be ΔU = q + w = -700 J + (-300 J) = -1000 J. This means the system loses 1000 joules of energy, which is consistent with the first law of thermodynamics stating that energy is conserved and transferred from one form to another.

What does it mean when the surroundings gain 250 joules of heat energy in relation to the system?

If the surroundings gain 250 joules of heat energy, it implies that the system has released that amount of heat energy to the surroundings. Therefore, for the system, q would be -250 J (negative because heat is leaving the system).

How is work performed by the surroundings different from work done on the system?

When work is performed by the surroundings, it means that the surroundings are doing work, losing energy, and transferring that energy to the system, making w positive for the system. Conversely, when work is done on the system, it gains energy, and the surroundings lose energy.

What is the formula used to calculate the work done by a gas during expansion or compression?

The work done by a gas during expansion or compression is calculated using the formula w = -pΔV, where p is the constant external pressure, and ΔV is the change in volume (final volume - initial volume). The negative sign indicates that work done by the gas (expansion) is considered negative, while work done on the gas (compression) is positive.

How can you convert work done in liters-atm to joules?

To convert work done from liters-atm to joules, use the conversion factor where 1 liter-atm equals 101.3 joules. Multiply the work in liters-atm by this factor to get the work in joules.

What happens to the internal energy of a gas when it expands against a constant external pressure?

When a gas expands against a constant external pressure, it does work on the surroundings, which results in a decrease in the gas's internal energy. This is because the gas is converting its internal energy into work done on the surroundings.

If 500 joules of heat energy is absorbed by a gas and it expands from 30 liters to 70 liters against a constant pressure of 2.8 atm, how much is the change in internal energy?

The change in internal energy (ΔU) can be calculated by adding the heat absorbed (q = +500 J) to the work done by the gas (w = -pΔV). The work done by the gas during expansion is w = -2.8 atm * (70 L - 30 L) = -112 L*atm. Converting this to joules gives w = -112 * 101.3 J/L*atm = -11,345.6 J. Therefore, ΔU = 500 J + (-11,345.6 J) = -10,845.6 J.

Outlines

00:00

🔍 Understanding Internal Energy Changes

This paragraph introduces the concept of internal energy changes in a system, focusing on the relationship between heat (q), work (w), and the change in internal energy (ΔU). It explains the first law of thermodynamics in the context of chemistry and physics, highlighting the difference in the sign conventions for q and w. The narrator walks through a problem where the system absorbs 300 joules of heat and has 400 joules of work done on it, resulting in an increase of 700 joules in internal energy. The explanation includes a visual representation of energy transfer between the system and its surroundings, illustrating the endothermic and exothermic processes.

05:00

🌡️ Calculating Energy Transfers in Systems

The second paragraph delves into calculating the change in internal energy when the system releases heat and does work. It presents a scenario where the system releases 700 joules of heat and does 300 joules of work, resulting in a net loss of 1000 joules of energy. The explanation includes the first law of thermodynamics, emphasizing energy transfer rather than creation or destruction. The narrator then introduces a new problem involving the surroundings gaining heat energy and work performed by the surroundings, leading to a calculation of ΔU as 220 joules.

10:00

📚 Work Done by a Gas Under Pressure

This paragraph discusses the work done by a gas when it expands or is compressed, starting with a visual representation of a gas in a cylinder. It explains the relationship between pressure, force, and volume change, and how work is calculated as force times displacement. The formula for work done on a gas is derived, highlighting the sign conventions for work during compression and expansion. The paragraph emphasizes the concept of energy storage in gases through pressure and the role of work in energy transfer, with a focus on the potential energy stored in high-pressure gases.

15:03

📉 Work and Energy Conversion in Gas Expansion

The fourth paragraph focuses on calculating the work done by a gas as it expands against a constant external pressure. It provides the formula for work (w = -pΔV) and applies it to a scenario where a gas expands from 25 liters to 40 liters against a pressure of 2.5 atm. The calculation results in a negative work value of 37.5 liters*atm, indicating energy transfer from the gas to the surroundings. The paragraph also includes a conversion of work from liters*atm to joules, resulting in -3790 joules of work done by the gas.

20:05

🔋 Internal Energy Change with Heat Absorption and Expansion

The final paragraph combines the concepts of heat absorption and work done to calculate the change in internal energy of a gas. It presents a problem where 500 joules of heat are absorbed, and the gas expands against a constant pressure of 2.8 atm from 30 liters to 70 liters. The calculation of work done by the gas during expansion is detailed, resulting in -11,345.6 joules. The change in internal energy (ΔU) is then calculated by adding the heat absorbed (q = +500 joules) to the work done (w = -11,345.6 joules), yielding a net decrease in internal energy of -10,845.6 joules.

Keywords

💡Internal Energy

Internal energy refers to the total energy contained within a system, which includes kinetic and potential energy of the particles. In the video, the concept is central as it discusses how changes in internal energy occur through heat transfer and work done on or by the system. The script uses the formula ΔU = Q + W to calculate changes in internal energy, where ΔU represents the change in internal energy, Q is the heat added to the system, and W is the work done on the system.

💡Heat Energy (q)

Heat energy, denoted as 'q', is the energy transferred between systems due to a temperature difference. In the video, it is explained that when a system absorbs heat (endothermic process), q is positive, and when it releases heat (exothermic process), q is negative. The script provides examples where 300 joules of heat energy is absorbed by the system, indicating an increase in internal energy.

💡Work (w)

Work, represented by 'w', is the energy transferred to or from a system by the application of force over a distance. The video clarifies that work done on a system increases its internal energy (w is positive), while work done by the system decreases it (w is negative). An example from the script is where 400 joules of work is done on the system, contributing to an increase in internal energy.

💡First Law of Thermodynamics

The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed, only transferred from one form to another. The video script emphasizes this by explaining how the change in internal energy (ΔU) is the sum of heat added to the system (Q) and work done on the system (W). This law is fundamental to understanding the examples and calculations presented in the video.

💡Endothermic Process

An endothermic process is a process where a system absorbs heat from its surroundings. The video script mentions this in the context of q being positive when heat is absorbed by the system, leading to an increase in internal energy. This is exemplified when the system absorbs 300 joules of heat energy, indicating an endothermic process.

💡Exothermic Process

An exothermic process is one where a system releases heat to its surroundings. In the video, this is discussed in relation to q being negative when heat is released by the system, resulting in a decrease in internal energy. The script provides an example where the system releases 700 joules of heat energy, illustrating an exothermic process.

💡Surroundings

In thermodynamics, the 'surroundings' refers to the environment outside of the system under study. The video script uses the term to describe the exchange of energy with the system, such as when the surroundings gain or lose heat energy. The concept is crucial for understanding the transfer of energy and how it affects the internal energy of the system.

💡Pressure

Pressure, in the context of the video, is the force exerted per unit area and plays a significant role in the work done by or on a gas. The script discusses how work is done against a constant external pressure when a gas expands or is compressed, using the formula w = -pΔV. An example is given where a gas expands against a constant external pressure of 2.5 atm, calculating the work done by the gas.

💡Volume

Volume in the video refers to the amount of space occupied by a substance, particularly a gas. Changes in volume (ΔV) are crucial in calculating the work done by or on a gas. The script explains how the work done by a gas expanding from 25 liters to 40 liters against a constant external pressure can be calculated using the formula w = -pΔV.

💡Expansion

Expansion, as discussed in the video, is the process where a gas increases in volume. It is associated with work being done by the gas as it pushes against its surroundings or the external pressure. The script provides a calculation of the work done by a gas as it expands from an initial volume to a larger volume, resulting in a negative value for work, indicating energy transfer from the gas to the surroundings.

Highlights

Introduction to chemistry problems related to internal energy, heat, and work.

Delta U (change in internal energy) equals Q (heat) plus W (work) in chemistry.

Explanation of when Q (heat) is positive or negative based on the system absorbing or releasing energy.

Explanation of when W (work) is positive or negative based on work done on or by the system.

First example: Calculate the internal energy when 300J of heat is absorbed, and 400J of work is done on the system.

System gains 700J total energy from both absorbed heat and work done on it.

Explanation of energy transfer between the system and surroundings, showing how energy flows between them.

Second example: System releases 700J of heat, and 300J of work is done by the system, resulting in a loss of 1000J of internal energy.

Clarification of energy transfer between system and surroundings in the second example.

Third example: If surroundings gain 250J of heat, and 470J of work is done by surroundings, the system's internal energy increases by 220J.

Fourth example: System absorbs 300J of heat and does 550J of work on surroundings, leading to a net loss of 250J.

Explanation of work done during gas expansion or compression, including formula for work in terms of pressure and volume.

Detailed derivation of work formula and sign conventions for work during expansion and compression.

Fifth example: Calculating work performed by a gas expanding from 25 to 40 liters against a constant pressure of 2.5 atm.

Sixth example: Calculating work to compress gas from 50 to 35 liters at constant pressure of 8 atm.

Seventh example: Calculating internal energy change when 500J of heat is absorbed, and gas expands from 30 to 70 liters against a 2.8 atm pressure.

Transcripts

in this video we're going to focus on

chemistry problems related to internal

energy heat and work

so let's start with this one calculate

the change in the internal energy of a

system

if 300 joules of heat energy

is absorbed by the system

and if 400 joules of work is done on the

system

now i don't know if you saw a previous

video that i created on the first law of

thermodynamics

but if you haven't in chemistry

delta u the change in internal energy is

equal to q plus w

in physics it's q minus w

now

q is positive whenever heat is absorbed

by the system

so that's during an endothermic process

q is negative

whenever heat is released by the system

so that's during an exothermic process

w is positive

whenever work

is done

on a system

so the internal energy of the system

will go up

and w is negative

whenever work is done by the system

so this would decrease

the internal energy of the system

so those are a few things to keep in

mind throughout this video

now let's get back to this problem

what is the value of q and what is the

value of w

now notice that the system

absorbs 300 joules of heat energy

because it absorbs energy

q is positive

now notice that work is done on a system

when work is done on a system

w is positive

so work is going to be positive 400

joules

now using that equation delta u

is q plus w so we have 300 joules of

heat energy absorbed by the system

plus 400 joules of work is done on a

system

so both of these events

work to increase

the internal energy of the system

so the change in the internal energy is

700

so if you want to draw a picture

let's say inside the box

represents the system

and everything outside of that

is the surroundings

so the system

absorbs 300 joules of energy

so the system gains

300

which means the surroundings

loses

300 joules

so this process is endothermic for the

system but exothermic for the

surroundings

now work is done on a system

400 joules of work is done on the system

so the system gains 400 joules of energy

but the surroundings

loses

400 joules of energy through work

so we could say work is done on a system

but work is done

by the surroundings

let's try this one

the system releases 700 joules of heat

energy

and 300 joules of work is done by the

system

calculate the change

in the internal energy of the system

so let's start with a picture first

so this is going to be the

system and outside of that we have the

surroundings

so the system releases 700 joules of

heat energy

so 700 joules of heat energy transfers

out of the system

so this is going to be negative 700 for

the system because it lost that energy

but the surroundings

gain 700 joules of heat energy

now

300 joules of work is done by the system

if the system is doing work it's

expending energy to do that

so the system is going to lose another

300 joules of energy

but the surroundings

will gain that 300 joules of energy

so the surroundings gain a total of 1

000 joules

but the system loses a total

of a thousand joules

and so we have the first law of

thermodynamics energy is neither created

or destroyed is simply transferred from

one place to another

so delta u

i forgot the u

which is q plus w

it's going to be negative 700

plus negative 300

so the change in the internal energy of

the system is negative one thousand

so if the system loses a thousand joules

of energy the surroundings gain a

thousand joules of energy but this is

the answer to the problem

number three

what is the change in the internal

energy of the system if the surroundings

gain 250 joules of heat energy

and if 470 joules of work is performed

by this romans

so we need to visualize the transfer of

energy

so if the surroundings gain 250 joules

of energy

is that heat energy flowing into the

system or into the surroundings

that energy

is flowing into the surroundings

if it flows if the surroundings gain

that energy

so therefore we could say q with respect

to the system

is negative 250 joules

because heat energy is coming out of the

system going into the surroundings

now 470 joules of work is performed by

the surroundings

whenever something performs work it

loses energy to do so

for example

if you perform the work required to lift

weights you have to burn energy to do it

so if the system is doing work

the system is losing energy so energy is

transferring i mean if the surroundings

is doing work

the surroundings loses energy which

means energy is transferred from the

surroundings to the system

so we got 470 joules of work

leaving the surroundings going to the

system so w

is positive 470.

so delta u which is q plus w

it's negative 250

plus 470

so that's 220

so i'm going to clarify

if work is performed by the surroundings

the surroundings is losing energy due to

work

and that energy is going into the system

which means

work is being done on a system if it's

done by the surroundings

and anytime work is done on a system

w is positive

number four

what is the change in the internal

energy of the system

if the surroundings releases 300 joules

of heat energy

and if the system

does 550 joules of work on the

surroundings

so go ahead and try this problem

now if the surroundings releases 300

joules of heat energy what does that

mean

well that means the surroundings is

losing energy so the system is gaining

that energy

so if the system absorbs 300 joules of

heat energy q is positive 300.

now the system does 550 joules of work

on the surroundings

so work is being done

by the system but on a surroundings

anytime work is done by the system

work is negative

energy is flowing out of the system

to the surroundings

so whenever work is done by something

energy is being consumed

so if work is done by the system the

system loses energy as

it's doing work it's expended energy and

if work is done on its surroundings the

surroundings is gaining energy so make

sure you understand what these

expressions mean so now let's calculate

the change

in the internal energy of the system

so it's q plus w

so the system gains 300 joules

as you can see that's flown into the

system but it's losing

550 joules

you can see that's flowing out of the

system

so then that result is that the system

is losing 250 joules of energy so delta

u is negative

we got a net energy flow

out of the system

number five

how much work is performed by a gas

as it expands from 25 liters to 40

liters

against a constant external pressure

of 2.5 atm

so what equation should we use for a

problem like this

well let's draw a picture

so this time the system

is a gas

so let's say this

cylinder is filled with

gas particles

so that's the system

and outside we have

the surroundings

now the surroundings

exerts a force on this gas

anytime you have a pressure

there's a force pressure is force

divided by area

so there's a constant external pressure

of 2.5 atm and that pressure exerts a

force in the gas as a result

the gas is going to

compress

and this problem is going to expand but

we'll talk about that later

right now i'm just deriving the formula

so as we apply a force on a gas

the gas will compress

so the volume is being reduced

and notice the change in the height of

the cylinder that's delta h

now to calculate the work done

by a gas

or on a gas

in this case it's on a gas

it's force

times the displacement

which in this case the displacement in

the y direction is the change in height

and based on this equation if you

rearrange it if you multiply both sides

by a

force

is pressure times area

now the volume of a cylinder is the area

times the change or times the height

so area times height is the volume of

the cylinder

the area

is the area of this circle

which is pi r squared

so this is the volume of the cylinders

pi r squared times height

so if the volume is area times height

then area times the change in height

must be the change in volume

now we need to add a negative sign to

make this work

due to the sign conventions

now

during compression

as was the case

relating to the picture that we have

the change in volume is negative the

volume is decreasing

and

a force was being applied on the gas in

order to compress it

so the surroundings was doing work on

the gas and

whenever work is done on the gas or on

the system w is positive

during expansion

whenever a gas expands

the volume increases

and the gas is doing work on the

surroundings as it expands

and so work

is going to be negative

so during compression

let me see if i could

draw this picture here

you're applying a force

to compress a gas

so you're doing work

on the gas

which means you're increasing the

internal energy of the gas

whenever the volume decreases

the pressure increases

based on boyle's law

so as you apply a force to compress a

gas what you're really doing is

you're expanding energy

and

that energy that you're expending it's

being stored

in the form of pressure so whenever you

compress a gas

you're transferring energy to that gas

you're increasing its pressure

and whenever that gas

decides to expand

it's going to apply an upward force and

as it applies in upward force

it can do work on the surroundings

so as the volume expands the pressure

reduces

so in order to store energy

in a gas you need to compress the gas

and to release that energy the gas has

to expand

and keep in mind the energy is stored in

the form of pressure which is a type of

potential energy

so gases that have a high pressure has a

lot of stored energy

gas is at low pressure

doesn't have much stored energy

so keep that in mind

so if you're doing work on the gas you

got to apply a force

so energy is being transferred from you

to the gas

now as the gas

expands against the surroundings energy

is flowing from the gas to the

surroundings

so during compression

w is positive work is done

on the gas

during expansion

w is negative work is done by the gas

and so the internal energy decreases

during expansion

but it increases

during express compression

now notice that delta v and w

always have opposite signs

so based on that

w is negative p delta v

so

when delta v is negative you're going to

have two negative signs

which means w has to be positive

as in the case of this problem

or

during expansion

when

delta v is positive

w has to be negative because a negative

times a positive number equals a

negative number

and so because these two signs are

opposite that's why we have the negative

sign in front of the equation

so just make sure you understand that

now let's focus on a problem at hand

how much work is performed

by a gas as it expands

from 25 liters to 40 liters

against a constant external pressure

of 2.5 atm

so we said the equation is negative p

delta v

the change in volume is the final volume

minus the initial volume

now this pressure

is not the internal pressure of the gas

because

that gradually changes from 4 atm to 2.5

atm

p

represents the constant external

pressure

that the gas has to work against

which was the 2.5 atm

the final volume is 40

the initial volume is 25

so during expansion delta v is positive

and we said w has to be negative

whenever a gas expands

so 40 minus 15 i mean 40 minus 25 is 15

and

2.5 times 15

that's 37.5

so the work is negative 37.5

liters times atm because the volume was

in liters and the pressure was an atm

so anytime a gas expands

the work done

by the gas is negative

now you need to be able to convert this

answer to joules

and here's the conversion that you need

one liter times one atm

is equal to 101.3 joules

so w

is three thousand

seven hundred

and ninety nine joules if you round it

to a nearest whole number

and don't forget this is negative

number six

how much work is required to compress a

gas from

50 liters to 35 liters

at a constant pressure

of 8 atm

so

during compression is the work going to

be positive or is it going to be

negative

to compress a gas

the work done on a gas is positive the

internal energy of the gas will increase

so let's go ahead and use this formula

it's negative p

delta v

so the pressure is constant it's 8 atm

and delta v is the final volume which is

35 liters

minus the initial

volume of 50 liters

so what we have is negative 8 atm

multiplied by a change in volume of

negative 15 liters

so therefore

the work required is positive

120

liters times atm

so now let's convert this to joules

so recall that one liter times one atm

is equal to 101.3 joules

so these units cancel

so it's 120 times 101.3

and you should get

12

156 joules

so that's the work required to compress

the gas from 50 liters to 35 liters at a

constant pressure of 8 atm

number seven

500 joules of heat energy was absorbed

from the surroundings

and the gas expanded from 30 liters to

70 liters

against the constant pressure of 2.8

atm calculate the internal energy change

in joules

so let's start with

this equation w is negative p

delta v

the change in volume is equal to the

final volume minus the initial volume

now the pressure

is 2.8 atm

and the volume

the final volume is 70 liters

minus the initial volume of 30 liters

so 70 minus 30

is 40 so we have negative 2.8 atm

multiplied by positive 40 liters

so the change in volume is positive due

to the expansion of the gas which means

work has to be negative

so negative 2.8 times 40

is negative 112

with the units being liters times atm

in the last problem w was 120

and that was liters times atm i didn't

convert it to joules but

you know how to convert it to joules in

this problem we need to because we want

the final answer to be in joules

so i'm going to multiply

this answer by 101.3 joules

per liter per atm

so that these units will cancel

so it's negative 112

times 101.3

so the work

due to the expansion of this gas

is negative 11

345.6 joules

so now

we need to calculate delta u

what is q is q

positive 500 or negative 500.

so notice that heat energy was absorbed

from the surroundings

if it's absorbed from the surroundings

that means heat flows from the

surroundings to the system

so heat energy is

absorbed

by the system so q is positive 500 the

system gains 500 joules from the

surroundings

the system being the gas by the way

so now delta u

is q plus w

so that's 500

plus

negative 11

345.6

so the change

in the internal energy of the system

is negative 10

845.6 joules

so this is the answer

you