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Chemical Thermodynamics - First Law of Thermodynamics

Grade 11ICSEChemistry

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

🔑Concepts

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Thermodynamics is the branch of science that deals with the relationship between heat and other forms of energy.

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A System is the part of the universe under observation, while the rest of the universe is the Surroundings. Systems are classified as Open (exchange mass and energy), Closed (exchange energy only), or Isolated (exchange neither).

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State Functions are properties that depend only on the initial and final states of the system, not the path taken (e.g., UU, HH, PP, VV, TT). Path Functions depend on the path (e.g., qq, ww).

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Internal Energy (UU) is the sum of all microscopic forms of energy (kinetic, potential, chemical) of a system.

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First Law of Thermodynamics (Law of Conservation of Energy): Energy can neither be created nor destroyed; it can only be converted from one form to another. Mathematically: ΔU=q+w\Delta U = q + w.

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Sign Convention (IUPAC): Heat absorbed by the system is +q+q; heat released is −q-q. Work done ON the system is +w+w; work done BY the system is −w-w.

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Enthalpy (HH) is defined as the total heat content of a system at constant pressure: H=U+PVH = U + PV.

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Extensive properties depend on the amount of matter (e.g., mass, volume, enthalpy), while Intensive properties are independent of the amount (e.g., temperature, density, molar heat capacity).

📐Formulae

ΔU=q+w\Delta U = q + w

w=−PextΔVw = -P_{ext} \Delta V

wrev=−2.303nRTlog⁡V2V1w_{rev} = -2.303 nRT \log \frac{V_2}{V_1}

ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

Cp−Cv=RC_p - C_v = R

q=mcΔTq = m c \Delta T

💡Examples

Problem 1:

A system absorbs 500 J500\text{ J} of heat and does 200 J200\text{ J} of work on the surroundings. Calculate the change in internal energy (ΔU\Delta U).

Solution:

Given: q=+500 Jq = +500\text{ J} (absorbed), w=−200 Jw = -200\text{ J} (done by the system). Using ΔU=q+w\Delta U = q + w, ΔU=500+(−200)=300 J\Delta U = 500 + (-200) = 300\text{ J}.

Explanation:

According to the First Law of Thermodynamics, the change in internal energy is the sum of heat added and work done. Since work is done by the system, it carries a negative sign.

Problem 2:

Calculate the work done when 22 moles of an ideal gas expand isothermally and reversibly from a volume of 10 L10\text{ L} to 20 L20\text{ L} at 300 K300\text{ K}. (Given R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1})

Solution:

w=−2.303nRTlog⁡V2V1w = -2.303 nRT \log \frac{V_2}{V_1}. Substituting values: w=−2.303×2×8.314×300×log⁡2010=−2.303×2×8.314×300×0.3010≈−3457.9 Jw = -2.303 \times 2 \times 8.314 \times 300 \times \log \frac{20}{10} = -2.303 \times 2 \times 8.314 \times 300 \times 0.3010 \approx -3457.9\text{ J}.

Explanation:

The formula for reversible isothermal expansion is used. The negative sign indicates work is done by the gas during expansion.

Problem 3:

For the reaction N2(g)+3H2(g)→2NH3(g)N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}, find the relationship between ΔH\Delta H and ΔU\Delta U at temperature TT.

Solution:

Δng=moles of gaseous products−moles of gaseous reactants=2−(1+3)=−2\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 2 - (1 + 3) = -2. Using ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT, we get ΔH=ΔU−2RT\Delta H = \Delta U - 2RT.

Explanation:

The change in enthalpy depends on the change in the number of gaseous moles. Here, since Δng\Delta n_g is negative, ΔH\Delta H will be less than ΔU\Delta U.

First Law of Thermodynamics - Revision Notes & Key Formulas | ICSE Class 11 Chemistry