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The Particulate Nature of Matter - Thermodynamic Systems

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The kinetic theory of matter states that all matter consists of tiny particles (atomsatoms or moleculesmolecules) in constant random motion.

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Internal Energy (UU) is defined as the total random kinetic energy (EkE_k) plus the total intermolecular potential energy (EpE_p) of the particles in a system: U=∑Ek+∑EpU = \sum E_k + \sum E_p.

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Temperature (TT) is a macroscopic measure of the average random kinetic energy per molecule (ar{E}_k). In an ideal gas, T∝EˉkT \propto \bar{E}_k.

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Heat (QQ) is the transfer of energy between a system and its surroundings due to a temperature difference. Energy always flows from higher temperature to lower temperature until thermal equilibrium is reached.

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Specific Heat Capacity (cc) is the amount of energy required to raise the temperature of 1 kg1 \text{ kg} of a substance by 1 K1 \text{ K} (or 1 ∘C1 \text{ } ^\circ C).

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Specific Latent Heat (LL) is the energy required to change the phase of 1 kg1 \text{ kg} of a substance at a constant temperature. LfL_f refers to the latent heat of fusion (solid-liquid), and LvL_v refers to the latent heat of vaporization (liquid-gas).

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During a phase change, the thermal energy added is used to increase the intermolecular potential energy (breaking or weakening bonds), while the average kinetic energy (and thus temperature) remains constant.

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The Mole (nn) is the amount of substance that contains NA=6.02×1023N_A = 6.02 \times 10^{23} particles (Avogadro's constant).

📐Formulae

Q=mcΔTQ = mc\Delta T

Q=mLQ = mL

n=NNAn = \frac{N}{N_A}

n=mMn = \frac{m}{M}

T(K)=T(∘C)+273.15T(K) = T(^\circ C) + 273.15

💡Examples

Problem 1:

Calculate the thermal energy required to heat 0.25 kg0.25 \text{ kg} of liquid water from 20 ∘C20 \text{ } ^\circ C to its boiling point of 100 ∘C100 \text{ } ^\circ C. The specific heat capacity of water is 4186 J kg−1K−14186 \text{ J kg}^{-1}\text{K}^{-1}.

Solution:

Using Q=mcΔTQ = mc\Delta T: \ m=0.25 kgm = 0.25 \text{ kg} \ c=4186 J kg−1K−1c = 4186 \text{ J kg}^{-1}\text{K}^{-1} \ ΔT=100−20=80 K\Delta T = 100 - 20 = 80 \text{ K} \ Q=0.25×4186×80=83,720 JQ = 0.25 \times 4186 \times 80 = 83,720 \text{ J}

Explanation:

The energy required is calculated by multiplying the mass, the specific heat capacity, and the change in temperature. Since the substance remains in the liquid phase throughout, we do not need to account for latent heat.

Problem 2:

How much energy is needed to completely melt 2.0 kg2.0 \text{ kg} of ice at 0 ∘C0 \text{ } ^\circ C? The specific latent heat of fusion of ice is 3.34×105 J kg−13.34 \times 10^5 \text{ J kg}^{-1}.

Solution:

Using Q=mLfQ = mL_f: \ m=2.0 kgm = 2.0 \text{ kg} \ Lf=3.34×105 J kg−1L_f = 3.34 \times 10^5 \text{ J kg}^{-1} \ Q=2.0×3.34×105=6.68×105 JQ = 2.0 \times 3.34 \times 10^5 = 6.68 \times 10^5 \text{ J}

Explanation:

Because the ice is already at its melting point, all added energy goes into changing the phase (increasing potential energy) rather than increasing the temperature.

Problem 3:

A sample contains 1.20×10241.20 \times 10^{24} molecules of H2OH_2O. Determine the number of moles present in this sample.

Solution:

n=NNAn = \frac{N}{N_A} \ N=1.20×1024N = 1.20 \times 10^{24} \ NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23} \text{ mol}^{-1} \ n=1.20×10246.02×1023≈1.99 moln = \frac{1.20 \times 10^{24}}{6.02 \times 10^{23}} \approx 1.99 \text{ mol}

Explanation:

The number of moles is the ratio of the total number of particles to the number of particles in one mole (Avogadro's constant).

Thermodynamic Systems - Revision Notes & Key Formulas | IB Grade 11 Physics