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Thermodynamic Systems (HL) (HL IB Physics)

Revision Note

Katie M

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Katie M

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Change in Internal Energy

  • The change in the internal energy increment U of an object is intrinsically related to a change in its temperature increment T

increment U space proportional to space increment T

  • When a container containing gas molecules is heated up, the molecules begin to move around faster, increasing their kinetic energy
    • In a solid, where the molecules are tightly packed, molecules begin to vibrate more as they are heated
    • Molecules in liquids and solids have both kinetic and potential energy because they are close together and bound by intermolecular forces
  • However, the molecules in an ideal gas are assumed to have no intermolecular forces
    • This means they do not possess potential energy, only kinetic energy

Change in internal energy, downloadable AS & A Level Physics revision notes

As the container is heated up, the gas molecules move faster with higher kinetic energy and therefore higher internal energy

  • The (change in) internal energy of an ideal gas is equal to:

increment U space equals space 3 over 2 N k subscript B increment T

  • Where
    • increment U = change in internal energy (J)
    • k subscript B = Boltzmann constant
    • increment T = change in temperature (K)
    • N = number of particles
  • Another form of this equation related to the translational kinetic energy of the particles is

increment U space equals space 3 over 2 n R increment T

  • Where:
    • n = number of moles of gas (mol)
    • R = molar gas constant

Worked example

A student suggests that, when an ideal gas is heated from 50°C to 150°C, the internal energy of the gas is tripled.

State and explain whether the student’s suggestion is correct.

Answer:

  • The change in internal energy of an ideal gas is directly proportional to its change in temperature

increment U space proportional to space increment T

  • The temperature change is the thermodynamic temperature i.e. Kelvin
  • The temperature change in degrees (from 50°C to 150°C) increases by three times
  • The temperature change in Kelvin is:

50°C + 273.15 = 323.15 K

150°C + 273.15 = 423.15 K

fraction numerator 423.15 over denominator 323.15 end fraction space equals space 1.3

  • The temperature change, in Kelvin, does not increase by three times, therefore, neither does the internal energy
  • Hence, the student is incorrect

Worked example

An ideal gas expands at constant pressure. The following data are available:

amount of gas = 126 mol

initial temperature of gas = −23.0°C

final temperature of gas = +27.0°C

Determine the change in internal energy of the gas during this expansion.

Answer:

  • The change in internal energy of a gas is equal to

increment U space equals space 3 over 2 n R increment T

  • Where
    • Amount of gas, n = 126 mol
    • Gas constant, R = 8.31 J mol−1 K−1
    • Change in temperature, increment T = 27 − (−23) = 50°C

increment U space equals space 3 over 2 cross times 126 cross times 8.31 cross times 50

increment U space equals space 7.85 space cross times space 10 to the power of 4 space straight J

Examiner Tip

If an exam question about an ideal gas asks for the total internal energy, remember that this is equal to the total kinetic energy since an ideal gas has zero potential energy

Work Done by a Gas

  • When a gas expands, it does work on its surroundings by exerting pressure on the walls of the container it's in
  • For a gas inside a piston, the force exerted by the gas pushes the piston outwards
    • As a result, work is done on the piston by the gas
  • Alternatively, if an external force is applied to the piston, the gas will be compressed
    • In this case, work is done on the gas by the piston

Work done by gas, downloadable AS & A Level Physics revision notes

The expansion of the gas does work on the piston by exerting a force over a distance, s

  • Assuming the volume of gas is kept at constant pressure, this means the force F exerted by the gas on the piston is equal to:

space p space equals space F over A space space space space space space rightwards double arrow space space space space space F space equals space p A

  • Where A = cross-sectional area of the cylinder (m2)
  • The definition of work done is:

W space equals space F s

  • Where s = displacement in the direction of force (m)
  • The displacement s of the gas multiplied by the cross-sectional area A is equal to the increase in volume ΔV of the gas:

W space equals space p A s

  • This gives the equation for the work done when the volume of a gas changes at constant pressure:

W space equals space p increment V

  • Where:
    • W = work done (J)
    • p = pressure of the gas (Pa)
    • ΔV = change in the volume of the gas (m3)
  • This equation assumes that the surrounding pressure does not change as the gas expands
    • This is true if the gas is expanding against the pressure of the atmosphere, which changes very slowly

p-V diagrams

  • Pressure-volume (p-V) diagrams are often used to represent changes in the state of a gas in thermodynamic processes

2-4-1-positive-and-negative-work-done-convention

Positive or negative work done depends on whether the gas is compressed or expanded

  • When a gas expands (at constant pressure) work done is positive
    • Volume increases +ΔV
    • Work is done by the gas +W

2-4-1-area-under-a-pv-diagram-gas-expansion

  • When a gas is compressed (at constant pressure) work done W is negative
    • Volume decreases −ΔV
    • Work is done on the gas −W

2-4-1-area-under-a-pv-diagram-gas-compression

  • When both the volume and pressure of gas changes

The work done can be determined from the area under a p-V diagram

2-4-1-area-under-a-pv-diagram-changing-pressure-and-volume

Worked example

When a balloon is inflated, its rubber walls push against the air around it.

Calculate the work done when the balloon is blown up from 0.015 m3 to 0.030 m3.

Atmospheric pressure = 1.0 × 105 Pa.

Answer:

  • The work done by a gas is equal to

W space equals space p increment V

  • Where the change in volume is

ΔV = final volume − initial volume = 0.030 − 0.015 = 0.015 m3

  • Therefore, work done is

W = (1.0 × 105) × 0.015 = 1500 J

Worked example

An ideal gas is compressed, as shown on the graph below. 

2-4-1-area-under-a-pv-diagram-worked-example2-4-1-area-under-a-pv-diagram-worked-example

(a)
For this change, state and explain whether work is done on the gas or by the gas
(b)
Determine the value of the work done and state whether it is positive or negative
 

Answer:

(a)

  • The volume decreases, therefore, work is done on the gas

(b)

  • The work done is equal to the area under the p-V diagram

2-4-1-area-under-a-pv-diagram-worked-example-ma

W space equals space 1 half open parentheses 600 cross times 10 cubed close parentheses open parentheses 400 cross times 10 to the power of negative 6 end exponent close parentheses space plus space open parentheses 300 cross times 10 cubed close parentheses open parentheses 400 cross times 10 to the power of negative 6 end exponent close parentheses

W = −240 J

  • When volume decreases, the work done is negative

Examiner Tip

The pressure p in the work done by a gas equation is not the pressure of the gas but the pressure of the surroundings. This is because when a gas expands, it does work on the surroundings.

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Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.