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AQA A-Level Physics - Engineering: Thermodynamics and Engines Summary

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A summary for the Thermodynamics and Engines subtopic (from the Engineering optional topic) in the AQA A-Level Physics Course.

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Subido en
27 de junio de 2023
Número de páginas
22
Escrito en
2022/2023
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Resumen

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ENGINEERING - THERMODYNAMICS AND ENGINES

THERMODYNAMICS:

First Law of Thermodynamics:
Q = ΔU + W
Energy transferred to the system = Change in internal energy of the system + work done by
the system

Example: A gas confined in a cylinder with a piston that can push down onto the gas.




If 100J energy is transferred to the system, and 80J of energy is transferred to the internal
energy of the gas (the kinetic energy of the particles increases), the work done by the
system is 20J.

Adibatic Processes:
- Q=0 ⇒ There is no net gain or loss of heat.
- ΔU + W = 0 ⇒ ΔU = -W ⇒ Any change in the internal energy of the system is caused
by work done
● Work done by the system: W is positive so the change internal energy is
negative so the internal energy of the system decreases
● Work done on the system: W is negative so the change is internal energy is
positive so the internal energy of the system increases
- In the piston and container: If the piston is pushed down and the air is compressed
adiabatically, work is done on the system, so W is negative and so the internal
energy of the system increases, causing the kinetic energy of the gas molecules to
increase and the temperature of the system to increase.
- pVγ = constant for adiabatic processes where γ = the adiabatic constant (= 5/3 for
gases typically)

Isothermal Processes:
- ΔT = 0 ⇒ Isothermal processes happen at a constant temperature.
- Q = ΔU + W. Constant temperature ⇒ no change in the internal energy of the gas, so
ΔU = 0 and so Q = W.
- At constant temperatures: pV = nRT = constant ⇒ p1V1 = p2V2

,Work done at constant pressure:
W = pΔV




Example: A piston moving upwards at a constant pressure has moved a distance Δh
W = Fd = FΔh
p = F/A ⇒ F = pA where A = the cross-sectional area
So, W = pAΔh = pΔV → Equation valid for a process taking place at constant pressure

When volume is constant, W = 0 as there is a zero change in volume.
Q = ΔU + W → When W = 0, Q = ΔU ⇒ All the energy transferred to the system is
transferred into raising the internal energy (and so the kinetic energy of the gas molecules)
of the system.

At constant volume:
- pV = nRT where V, n (amount of substance in moles which is constant for a closed
system), and R are constant ⇒ p/T = nR/V = constant ⇒ p1/T1 = p2/T2

At constant pressure:
- pV = nRT where p, n and R are constant ⇒ V/T = nR/P = constant ⇒ V1/T1 = V2/T2

,
, First Law of Thermodynamics: The energy transferred to a system by heating is equal to
the sum of the increase in the internal energy and the work done by the system (Q = ΔU +
W). Q = Energy transferred to the system by heating / cooling, ΔU = increase in internal
energy of the system and W = work done by the system.

- Note: Q is the energy transferred to the system through heating, therefore if Q is
negative, energy is transferred away from the system through cooling.

- Note: W is the work done by the system and this occurs when the gas expands.
Therefore, if W is negative, work is being done on the system, and this occurs when
the gas is being compressed.

- Note: The internal energy of the system (U) is equal to the sum of all the kinetic
energies and potential energies of all the particles in the system. As ΔU represents
an increase in the internal energy of the system, if ΔU is negative, the internal energy
of the system decreases.

A system is a region containing a body of gas:
- Open system: Gas can flow in, out or through the system. Therefore, the gas can
cross the boundaries of the system. Example: Aerosol can.

- Closed system: No gas can leave or enter the system. The boundaries of the system
may change when the system changes volume. Example: Air inside a balloon.

Application of the first law of thermodynamics: The human metabolism. Q is negative as the
human body transfers heat to the surroundings, and W is positive as work is being done by
the body. In this situation, the internal energy of the human body must decrease as ΔU must
be negative to make the equation hold true:

-Q = ΔU + W ⇒ -ΔU = Q + W


Non-Flow Processes:
- Changes which occur in closed systems as the gas is not allowed to flow across a
boundary.
- To apply the first law of thermodynamics to non-flow processes, the gas in the
system is assumed to be ideal:
● Follows the gas laws perfectly at all temperatures - so there is no other
interaction between gas molecules other than perfectly elastic collisions. This
means there are no intermolecular forces between the molecules. As
potential energy is associated with intermolecular forces, an ideal gas has 0
potential energy ⇒ ΔU = sum of the kinetic energies of all its particles.
- As the gas is assumed to be ideal, pV = nRT (the ideal gas equation) can be used. [p
= pressure, V = volume, n = amount of substance in moles, R = gas constant and T =
temperature is K]
- As the system is closed in non-flow processes, n is constant ⇒ pV/T = constant ⇒
p1V1/T1= p2V2/T2
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