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# Chapter 15. Work, Heat, and the First Law of Thermodynamics

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Chapter 15. Work, Heat, and the First Law of Thermodynamics

Heat and Work in Ideal-Gas Processes
Consider a stationary cylinder of ideal gas sealed at one end by a moveable piston (which is not moving at this time). KEcyl =0 and PEcyl = 0 (assume cylinder position at y = 0). But since T > 0 kelvin, average KE per molecule = 3/2 kT Total KE = 3/2 NkT or 3/2 nRT. For a monatomic gas: KEtot = Uint = 3/2 nRT For non-monatomic gases: Uint is still proportional to T, but greater than that of a monatomic gas.

Heat and Work in Ideal-Gas Processes
How can we change the amount of internal energy in our system of an ideal gas? We have studied 2 energy transfer mechanisms, heat (Q) and work (W). Let’s look at both of these mechanisms.

Review: What is the best definition of heat?
the amount of thermal energy in an object. energy transfer from a hotter object  to a colder object. how high the temperature of an object is. all of the above Answer: B

Heat, Temperature, and Thermal Energy
Internal energy U is an energy of the system due to the   motion of its atoms and molecules. Any system has a   thermal energy even if it is isolated and not interacting   with its environment. The units of U are Joules. Heat Q is energy transferred between the system and the   environment as they interact due to a difference in temperature. The units of Q are Joules. Temperature T is a state variable that quantifies the   “hotness” or “coldness” of a system. Although proportional to the thermal energy of a system, it is not the same thing! A temperature  difference is required in order for heat to be transferred   between the system and the environment. The units of T   are degrees

If heat is the only energy transfer mechanism
ΔU = Uf -Ui = Q Q is positive when the system gains energy. This means that the environment has a higher temperature than the system. Q is negative when the system loses energy. This means that the environment has a lower temperature than the system.

Work done by the system W = |F| |Δs| cos θ
If the piston moves to the right: Work done by the gas molecules on the piston is positive (Fgas is to the right, piston moves to the right). energy is added to the piston, energy is taken away from the gas and the gas expands. If the piston moves to the left: Work done by the gas molecules on the piston is negative (Fgas is to the right, gas molecules move left). Energy is added to the gas and the gas compresses.

15.3 The First Law of Thermodynamics
The internal energy of a system changes due to heat and work: Heat is positive when the system gains heat and negative when the system loses heat. Work is positive when it is done by the system (gas expands) and negative when it is done on the system (gas compresses).

There’s not sufficient information to tell.
A gas cylinder and piston are covered with heavy insulation so there can be no heat exchange with the environment. The piston is pushed into the cylinder by an external force, compressing the gas. According to the 1st Law of Thermodynamics, the gas temperature: decreases. increases. doesn’t change. There’s not sufficient information to tell. Answer: B

insulation implies no heat exchange with environment
ΔU = Q – Wby insulation implies no heat exchange with environment Work done by gas is negative. Gas pushes left, piston compresses gas to the right: ΔU = – – Wby ΔU is proportional to temperature increase. Therefore, temperature increases. Answer: B

EOC # 1 The internal energy of a system changes because the system gains 165 J due to heat transfer and does 312 J of work on the environment. In returning to its initial state *, the system loses 114 J of heat. Determine the magnitude of the work done during the return process? Is the work done by the system, or on the system? * Initial state (for a gas) means same volume, pressure, temperature, and therefore internal energy.

EOC # 1-Answer The internal energy of a system changes because the system gains 165 J due to heat transfer and does 312 J of work. In returning to its initial state *, the system loses 114 J of heat. Determine the magnitude of the work done during the return process? 261 J Is the work done by the system, or on the system? Since the sign of Wby (work done by the system) was negative, Work must be done on the system.

15.3 The First Law of Thermodynamics
Example 2 An Ideal Gas The temperature of three moles of a monatomic ideal gas is reduced from 540K to 350K as 5500J of heat flows into the gas. Find (a) the change in internal energy and (b) the work done by the gas.

15.3 The First Law of Thermodynamics
(b)

Ideal Gas Processes isobaric: constant pressure
15.4 Thermal Processes Ideal Gas Processes isobaric: constant pressure isochoric: constant volume isothermal: constant temperature adiabatic: no transfer of heat

An isobaric process is one that occurs at constant pressure.
15.4 Thermal Processes An isobaric process is one that occurs at constant pressure. According to the 1st Law of Thermodynamics: ΔU = Q – Wby During an isobaric process: ΔU = Q – P ΔV: energy is transferred by both work and heat.

The PV diagram for an work done during an isobaric process is a horizontal line

An isobaric process During an isobaric process, a system gains 1500 J of heat. The internal energy of the system increases b y 4500 J and the volume decreases by 0.01 m3 . Find the pressure of the system.

An isobaric process During an isobaric process, a system gains 1500 J of heat. The internal energy of the system increases b y 4500 J and the volume decreases by 0.01 m3 . Find the pressure of the system. Answer: 3.0 x 105 Pa

The work done by the gas equals the area under the PV curve

The area under a pressure-volume graph is
15.4 Thermal Processes Example 4 Work and the Area Under a Pressure-Volume Graph Determine the work for the process in which the pressure, volume, and temp- erature of a gas are changed along the straight line in the figure. The area under a pressure-volume graph is the work for any kind of process.

Since the volume increases, the work is positive.
15.4 Thermal Processes Since the volume increases, the work is positive. Estimate that there are 9colored squares in the drawing.

Two process are shown that take an ideal gas from state 1 to state 3
Two process are shown that take an ideal gas from state 1 to state 3. Compare the work done by process A to the work done by process B. Answer: A WA > WB WA < WB WA = WB = 0 WA = WB but neither is zero

15.4 Thermal Processes isochoric: constant volume But W = PΔV = 0: during an isochoric process, energy is transferred by heat only For the isochoric process the area under the curve is equal to zero.

15.5 Thermal Processes Using and Ideal Gas
ISOTHERMAL EXPANSION OR COMPRESSION Isothermal expansion or compression of an ideal gas ΔU = Q – Wby But ΔU α ΔT If T does not change, ΔU = ΔT = 0! And Q = W

EXAMPLE The work of an isothermal compression
QUESTION: A cylinder contains 7.0 g of N2 gas. The gas compresses to half its volume at a constant temperature of 80˚C. How much work must be done by the gas? By how much does the internal energy of the gas change? How much heat was added or taken away from the gas?

EXAMPLE The work of an isothermal compression
Answer: A cylinder contains 7.0 g of N2 gas. The gas compresses to half its volume at a constant temperature of 80˚C. How much work must be done by the gas? J (work was done on the gas to compress it) By how much does the internal energy of the gas change? 0J How much heat was added or taken away from the gas? 508 J of heat was taken away from the gas. Q – Wby = 0

Focus on Concepts #9 Five moles of an ideal gas are compressed isothermally from A to B, as the graph illustrates. What is the work involved if the temperature of the gas is 320 K? Be sure to include the correct algebraic sign.

15.5 Thermal Processes Using and Ideal Gas
ADIABATIC EXPANSION/COMPRESSION According to the 1st Law of Thermodynamics: ΔU = Q – Wby but Q = 0, since walls are insulated ΔU = – Wby For a monatomic ideal gas: The red curve shows an adiabatic expansion of an ideal gas. The blue curves are isotherms at Ti and Tf. Adiabatic curves can be approximated as linear.

Adiabatic Processes without adiabatic (insulating) walls
Rapid expansion or compression does not allow the gas and surroundings to come to equilibrium bicycle pump air being forced to rise or sink due to atmospheric conditions loading/ unloading a pneumatic lift.

15.5 Thermal Processes Using and Ideal Gas
ADIABATIC EXPANSION OR COMPRESSION γ = 5/3 for a monatomic gas γ =7/5 for a diatomic gas (often used for air. which will be discussed in the following section

EOC #27 A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of 23° C is compressed adiabatically to a temperature of 710° C. Assume the air to be an ideal gas for which γ = 7 5. Find the compression ratio, which is the ratio of the initial volume to the final volume.

EOC #27 A diesel engine does not use spark plugs to ignite the fuel and air in the cylinders. Instead, the temperature required to ignite the fuel occurs because the pistons compress the air in the cylinders. Suppose that air at an initial temperature of 23° C is compressed adiabatically to a temperature of 710° C. Assume the air to be an ideal gas for which γ = 7 5. Find the compression ratio, which is the ratio of the initial volume to the final volume. Ans: 20.1

Molar specific heat capacity
For a solid or liquid: Q = cm ∆T, where c is the specific heat capacity in Joules/kg kelvin. For a gas, we use number of moles (n) instead of mass: Q = Cn ∆T where C is the molar specific heat capacity in Joules/mol kelvin.

Molar specific heat capacity for a monatomic ideal gas
At constant pressure Cp = 5/2R At constant volume Cv = 3/2R. Q = Cn ∆T ; so for the same amount of heat (Q) added to a gas, the gas at constant volume will show a greater temperature change because it is not losing energy by expanding. = 5/3

Molar specific heat capacity
Two containers hold equal masses of helium gas (monatomic) at equal temperature. You supply 10 J of heat to container A while not allowing the volume to change. You supply 10 J of heat to container B while not allowing the pressure to change. Is TfA greater than, less than or equal to TfB? Explain, using the 1st Law of Thermodynamics. TfA > TfB TfA < TfB TfA = TfB

EOC # 33 Heat, Q, is added to a monatomic ideal gas at constant pressure. As a result, work, W is done by the gas. What is Q/W, the ratio of heat added to work done by the gas? This is a numerical value, with no variables.

EOC # 33 Heat, Q, is added to a monatomic ideal gas at constant pressure. As a result, work, W is done by the gas. What is Q/W, the ratio of heat added to work done by the gas? This is a numerical value, with no variables. Q/W = 5/2/1 = 2.5.

EOC #34 Three moles of a monatomic ideal gas are heated at a constant volume of 2.60 m3. The amount of heat added is 4.76 x103 J. What is the change in the temperature of the gas? Find the change in its internal energy. Determine the change in pressure.

EOC #34 Three moles of a monatomic ideal gas are heated at a constant volume of 2.60 m3. The amount of heat added is 4.76 x103 J. What is the change in the temperature of the gas? K Find the change in its internal energy J Determine the change in pressure Pa

Which of the following processes involve heat (energy transfer due to ∆T)?
The brakes in your car get hot when you stop. You push a rigid cylinder of gas across a frictionless surface. A steel block is placed on top of a candle. You push a piston into a cylinder of gas, increasing the temperature of the gas. All of the above Answer: C

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