First Law of Thermodynamics (DP IB Physics)
Revision Note
The First Law of Thermodynamics
The first law of thermodynamics is based on the principle of conservation of energy
When energy is put into a gas by heating it or doing work on it, its internal energy must increase:
energy supplied by heating = change in internal energy + work done on the system
The first law of thermodynamics is therefore defined as:
Where:
Q = energy supplied to the system by heating (J)
ΔU = change in internal energy (J)
W = work done by the system (J)
The first law of thermodynamics applies to all situations, not just to gases
There is an important sign convention used for this equation
A positive value for internal energy (+ΔU) means:
The internal energy ΔU increases
Heat Q is added to the system (+Q)
Work W is done on the system (–W)
A negative value for internal energy (−ΔU) means:
The internal energy ΔU decreases
Heat Q is taken away from the system (–Q)
Work W is done by the system (+W)
Graphs of Constant Pressure & Volume
Graphs of pressure p against volume V can provide information about the work done and internal energy of the gas
The work done is represented by the area under the line
A constant pressure process is represented as a horizontal line
If the volume is increasing (expansion), work is done by the gas (on the surroundings) and internal energy decreases (ΔU = q − W)
If the arrow is reversed and the volume is decreasing (compression), work is done on the gas and internal energy increases (ΔU = q + W)
The volume of the gas is made smaller, so more collisions between the molecules of the gas and the walls of the container occur. This creates a higher pressure.
A constant volume process is represented as a vertical line
In a process with constant volume, the area under the curve is zero
Therefore, no work is done when the volume stays the same
Work is only done when the volume of a gas changes
Worked Example
The volume occupied by 1.00 mol of a liquid at 50°C is 2.4 × 10−5 m3. When the liquid is vaporised at an atmospheric pressure of 1.03 × 105 Pa, the vapour occupies a volume of 5.9 × 10−2 m3.
The latent heat to vaporise 1.00 mol of this liquid at 50°C at atmospheric pressure is 3.48 × 104 J.
For this change of state, determine the increase in internal energy ΔU of the system.
Answer:
Step 1: List the known quantities
Thermal energy, Q = 3.48 × 104 J
Atmospheric pressure, p = 1.03 × 105 Pa
Initial volume = 2.4 × 10−5 m3
Final volume = 5.9 × 10−2 m3
Step 2: Calculate the work done W
The work done by a gas at constant pressure is
Where the change in volume is:
ΔV = final volume − initial volume = (5.9 × 10−2) − (2.4 × 10−5) = 0.059 m3
Since the volume of the gas increases, the work done is positive
W = (1.03 × 105) × 0.059 = 6077 = 6.08 × 103 J
W = +6.08 × 103 J
Step 3: Substitute the values into the equation for the first law of thermodynamics
From the first law of thermodynamics:
ΔU = (3.48 × 104) − (6.08 × 103) = 28 720 J
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