Internal Energy, Heat, and Work Thermodynamics, Pressure & Volume, Chemistry Problems
Summary
TLDRThis 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
🔍 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.
🌡️ 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.
📚 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.
📉 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.
🔋 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.
Mindmap
Keywords
💡Internal Energy
💡Heat Energy (q)
💡Work (w)
💡First Law of Thermodynamics
💡Endothermic Process
💡Exothermic Process
💡Surroundings
💡Pressure
💡Volume
💡Expansion
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
Weitere ähnliche Videos ansehen
First Law of Thermodynamics, Basic Introduction - Internal Energy, Heat and Work - Chemistry
Thermodynamics - A-level Physics
Specific Heat Capacity
#8 Física 10º ano - Energia e movimentos 🏃♀️
First law of thermodynamics / internal energy | Thermodynamics | Physics | Khan Academy
12th Physics | Chapter 4 | Thermodynamics | Lecture 2 | Maharashtra Board |
5.0 / 5 (0 votes)