The Second law of thermodynamics IB DP Physics Study Notes - 2025 Syllabus
The Second law of thermodynamics IB DP Physics Study Notes
The Second law of thermodynamics IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
the second law of thermodynamics refers to the change in entropy of an isolated system and sets constraints on possible physical processes and the overall evolution of the system.
Processes in real isolated systems are almost always irreversible, and consequently, the entropy of a real isolated system always increases.
The entropy of a non-isolated system can decrease locally, but this is compensated by an equal or greater increase in the entropy of the surroundings.
Isovolumetric, isobaric, isothermal, and adiabatic processes are obtained by keeping one variable fixed.
Adiabatic processes in monatomic ideal gases can be modeled by the equation:
\( PV^{5/3} = \text{constant} \)
Standard level and higher level: There is no Standard level content
Additional higher level: 8 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
The 2nd law of thermodynamics:
Kelvin version:
It is impossible to extract energy from a hot reservoir and transfer this entirely into work
Clausius version:
It is impossible for energy to spontaneously transfer from a body at a lower temperature to one at higher temperature without doing work on the system.
In every process, the total entropy of any isolated system, or the Universe, always increases
The natural way of nature is for things to tend toward disorder.
Can the room become ordered again?
If yes, what would be the conditions for it to become more ordered?
As can be seen in the table, the more opportunities for the particles to move more randomly, the more disordered they will be, increasing entropy.
This is tied to statistics and the idea of multiplicity. The more possible states for the particles, the greater the amount of entropy.
Consequences of the 2nd law of thermodynamics
1. Heat cannot flow spontaneously from a cold object to a hot one.
This would create greater order in the particles, but that requires work to be done.
2. $\Delta S_{Universe} > 0$
It is possible to decrease entropy locally, but it must always be accompanied by an increase elsewhere. This is what happens when you make ice cubes in the freezer.
3. A closed reversible process must have $\Delta S = 0$
These are idealized systems. Real isolated reversible systems will always lose energy to the surroundings and have $\Delta S > 0$.
Microstates
- To better understand entropy, we need to consider the idea of microstates. Let’s start by considering 2 states, speed and direction.
- Each of the possible speeds would be a microstate. Likewise each of the possible directions would be a microstate.
- Since the particles here can only travel in one direction with one speed, they have few microstates and low entropy.
- In this example, the particles have more speed microstates and more directions microstates, meaning that they also have greater entropy.
Work done in an isothermal change
Isothermal = constant temperature (T does not Change )
For an ideal gas, internal energy equals the sum of mean kinetic energy of particles of gas (\(U = N \bar{E_k}\)).
We also know that the average kinetic energy is proportional to the temperature of the gas (\(\bar{E_k} \propto T\)).
So, if \(T\) is constant, Internal energy = constant.
For \(Q = \Delta U + W\) with \(\Delta U = 0 \Rightarrow Q = W\):
- If \(Q > 0\), work done by system on surroundings → expansion of gas does work.
- If \(Q < 0\), work done by surroundings on system → gas compresses due to work being done on it.
EXAMPLE:
A graduated syringe filled with air is placed in an ice bath and allowed to reach the temperature of the water.
Demonstrate that \( p_1V_1 = p_2V_2 \) (Boyle’s Law).
▶️Answer/Explanation
SOLUTION:
1. Record the initial state after equilibrium is reached:
\( p_1 = 15 \, \text{units} \)
\( V_1 = 10 \, \text{units} \)
\( T_1 = 0^\circ \text{C} \)
2. Record the final state after adjustments:
\( p_2 = 30 \, \text{units} \)
\( V_2 = 5 \, \text{units} \)
\( T_2 = 0^\circ \text{C} \)
3. Calculate and verify Boyle’s Law:
Initial state: \( p_1V_1 = 15 \cdot 10 = 150 \)
Final state: \( p_2V_2 = 30 \cdot 5 = 150 \)
Conclusion:
Since \( p_1V_1 = p_2V_2 \), Boyle’s Law is verified. This confirms that, at a constant temperature, the product of pressure and volume for a given amount of gas remains constant.
Work done in an isovolumetric change
Isovolumetric = constant volume
- When V = constant then NO work can be done (Why?)
- For $Q = \Delta U + W$ with $W = 0 \rightarrow Q = \Delta U$
- If $Q > 0$, there is an increase of internal energy of system → temperature goes up
- If $Q < 0$, there is a decrease of internal energy of system → temperature goes down
In an isovolumetric process, V does not change.
- We have already seen an isovolumetric experiment when we studied the concept of absolute zero:
- During an isovolumetric process the temperature and the pressure change.
- Note that the volume was kept constant in this experiment.
EXAMPLE:
Show that for an isolated ideal gas \(p \propto T\) during an isovolumetric process.
▶️Answer/Explanation
SOLUTION: Use \(pV = nRT\). Then:
\[
p = \left(\frac{nR}{V}\right)T.
\]
Isolated means \(n\) is constant (no gas is added to or lost from the system).
Isovolumetric means that \(V\) is constant.
Then \(n\) and \(V\) are constant (as is \(R\)). Thus:
\[
p = \left(\frac{nR}{V}\right)T = (\text{CONST})T
\]
\[
p \propto T. \quad (\text{isovolumetric process})
\]
FYI: Isovolumetric is sometimes called isochoric.
Work done in an adiabatic process
In an adiabatic process, no energy is supplied to the system. Instead, the work done directly results in changes to the internal energy.
- Adiabatic = no energy is transferred (\(Q = 0\))
- For \(Q = \Delta U + W\) with \(Q = 0 \Rightarrow \Delta U = -W\):
- If \(W > 0\), the temperature of the gas drops as work is done on the surroundings.
- If \(W < 0\), the temperature of the gas rises as work is done by the surroundings.
Without proof, there is a relationship between pressure and volume in an adiabatic process for a monatomic ideal gas
EXAMPLE:
The volume of an ideal monatomic gas is reduced in an adiabatic compression by a factor of 8.0. Determine the factor by which the pressure in the 1 gas changes
▶️Answer/Explanation
SOLUTION:
$P_1V_1^{\frac{5}{3}} = P_2V_2^{\frac{5}{3}} \Rightarrow \frac{P_2}{P_1} = \left(\frac{V_1}{V_2}\right)^{\frac{5}{3}} = 8^{\frac{5}{3}} = 32$