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The Particulate Nature of Matter - Thermodynamics

Grade 12IBPhysics

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

πŸ”‘Concepts

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The kinetic molecular theory states that matter is composed of small particles (atoms or molecules) in constant, random motion. The average kinetic energy of these particles is directly proportional to the absolute temperature of the substance in Kelvin (TT).

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Internal Energy (UU) is defined as the sum of the total random kinetic energy and the total intermolecular potential energy of all the particles in a system. For an ideal gas, there are no intermolecular forces, so the internal energy consists only of kinetic energy.

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Temperature must be used in Kelvin (KK) for all thermodynamic calculations. The conversion is T(K)=t(∘C)+273.15T(K) = t(^{\circ}C) + 273.15.

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Specific Heat Capacity (cc) is the amount of energy required to raise the temperature of 1 kg1\,kg of a substance by 1 K1\,K. During a temperature change, Q=mcΞ”TQ = mc\Delta T.

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Specific Latent Heat (LL) is the energy required to change the phase of 1 kg1\,kg of a substance at a constant temperature. LfL_f is for fusion (melting/freezing) and LvL_v is for vaporization (boiling/condensing). Formula: Q=mLQ = mL.

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An Ideal Gas is a theoretical gas that obeys the gas laws at all pressures, volumes, and temperatures. Assumptions include: particles are point masses, collisions are perfectly elastic, and there are no intermolecular forces.

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The equation of state for an ideal gas is PV=nRTPV = nRT or PV=NkBTPV = Nk_BT, where nn is the number of moles and NN is the number of molecules.

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The First Law of Thermodynamics is a statement of conservation of energy: Q=Ξ”U+WQ = \Delta U + W, where QQ is heat added to the system, Ξ”U\Delta U is the change in internal energy, and WW is the work done by the system.

πŸ“Formulae

Q=mcΞ”TQ = mc\Delta T

Q=mLQ = mL

PV=nRTPV = nRT

PV=NkBTPV = Nk_BT

n=NNA=mMn = \frac{N}{N_A} = \frac{m}{M}

Ekˉ=32kBT=32RNAT\bar{E_k} = \frac{3}{2}k_BT = \frac{3}{2}\frac{R}{N_A}T

W=PΞ”VW = P\Delta V

Ξ”U=Qβˆ’W\Delta U = Q - W

πŸ’‘Examples

Problem 1:

Calculate the energy required to convert 0.50 kg0.50\,kg of ice at 0∘C0^{\circ}C into water at 20∘C20^{\circ}C. (Given: Lf=3.34Γ—105 J kgβˆ’1L_f = 3.34 \times 10^5\,J\,kg^{-1} and cwater=4186 J kgβˆ’1Kβˆ’1c_{water} = 4186\,J\,kg^{-1}K^{-1})

Solution:

First, calculate energy for melting: Q1=mLf=0.50Γ—3.34Γ—105=1.67Γ—105 JQ_1 = mL_f = 0.50 \times 3.34 \times 10^5 = 1.67 \times 10^5\,J. Next, calculate energy to raise temperature: Q2=mcΞ”T=0.50Γ—4186Γ—(20βˆ’0)=4.186Γ—104 JQ_2 = mc\Delta T = 0.50 \times 4186 \times (20 - 0) = 4.186 \times 10^4\,J. Total energy Q=Q1+Q2=1.67Γ—105+0.4186Γ—105=2.0886Γ—105 JQ = Q_1 + Q_2 = 1.67 \times 10^5 + 0.4186 \times 10^5 = 2.0886 \times 10^5\,J.

Explanation:

The process involves two stages: a phase change (where temperature is constant) followed by a temperature increase of the liquid water.

Problem 2:

A gas cylinder contains 2.0 mol2.0\,mol of an ideal gas at a pressure of 1.5Γ—105 Pa1.5 \times 10^5\,Pa and a temperature of 300 K300\,K. If the volume is halved and the temperature is increased to 400 K400\,K, calculate the new pressure.

Solution:

Using the ideal gas law ratio: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}. Given V2=0.5V1V_2 = 0.5V_1, rearrange for P2P_2: P2=P1Γ—(V1V2)Γ—(T2T1)=1.5Γ—105Γ—(V10.5V1)Γ—(400300)=1.5Γ—105Γ—2Γ—1.333=4.0Γ—105 PaP_2 = P_1 \times (\frac{V_1}{V_2}) \times (\frac{T_2}{T_1}) = 1.5 \times 10^5 \times (\frac{V_1}{0.5V_1}) \times (\frac{400}{300}) = 1.5 \times 10^5 \times 2 \times 1.333 = 4.0 \times 10^5\,Pa.

Explanation:

Since the number of moles nn is constant, we can use the combined gas law to find the final pressure based on the changes in volume and temperature.

Thermodynamics - Revision Notes & Key Formulas | IB Grade 12 Physics