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Chemical Thermodynamics - First Law of Thermodynamics

Grade 11CBSEChemistry

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

🔑Concepts

Thermodynamics deals with the energy changes accompanying physical and chemical processes.

A System is the part of the universe under observation, while the Surroundings comprise the rest of the universe.

Types of Systems: Open (exchange of matter and energy), Closed (exchange of energy only), and Isolated (no exchange of matter or energy).

Internal Energy (UU): The total energy contained within a system. It is a state function and an extensive property.

First Law of Thermodynamics: 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.

Sign Convention (IUPAC): Heat absorbed by the system (qq) is positive (++); heat released is negative (-). Work done on the system (ww) is positive (++); work done by the system is negative (-).

Work (ww): For an irreversible process against constant external pressure, w=PextΔVw = -P_{ext} \Delta V. For a reversible isothermal process, wrev=2.303nRTlogV2V1w_{rev} = -2.303 nRT \log \frac{V_2}{V_1}.

Enthalpy (HH): The total heat content of a system at constant pressure, defined as H=U+PVH = U + PV. The change in enthalpy is given by ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V.

Relation between ΔH\Delta H and ΔU\Delta U: For reactions involving gases, ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT, where Δng\Delta n_g is the change in the number of moles of gaseous products and reactants.

Heat Capacity: The amount of heat required to raise the temperature of a substance by 1C1^{\circ}C (or 1K1 K). At constant volume, it is CvC_v, and at constant pressure, it is CpC_p. For one mole of an ideal gas, CpCv=RC_p - C_v = R.

📐Formulae

ΔU=q+w\Delta U = q + w

w=Pext(V2V1)w = -P_{ext}(V_2 - V_1)

wrev=2.303nRTlog10V2V1w_{rev} = -2.303 nRT \log_{10} \frac{V_2}{V_1}

H=U+PVH = U + PV

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

q=CΔT=mcΔTq = C \Delta T = m c \Delta T

CpCv=RC_p - C_v = R

💡Examples

Problem 1:

Calculate the work done when 2.02.0 moles of an ideal gas expand isothermally and reversibly from a volume of 10L10 L to 20L20 L at 300K300 K.

Solution:

Given: n=2.0moln = 2.0 mol, R=8.314JK1mol1R = 8.314 J K^{-1} mol^{-1}, T=300KT = 300 K, V1=10LV_1 = 10 L, V2=20LV_2 = 20 L. Using wrev=2.303nRTlogV2V1w_{rev} = -2.303 nRT \log \frac{V_2}{V_1}: w=2.303×2.0×8.314×300×log2010w = -2.303 \times 2.0 \times 8.314 \times 300 \times \log \frac{20}{10} w=2.303×2.0×8.314×300×0.3010w = -2.303 \times 2.0 \times 8.314 \times 300 \times 0.3010 w3455.3Jw \approx -3455.3 J

Explanation:

Since the gas is expanding, the work is done by the system, which results in a negative value for ww according to IUPAC convention.

Problem 2:

For the reaction N2(g)+3H2(g)2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g), the value of ΔH\Delta H is 92.38kJ-92.38 kJ at 298K298 K. Calculate ΔU\Delta U for the reaction.

Solution:

Reaction: N2(g)+3H2(g)2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g). Δng=nproducts(g)nreactants(g)=2(1+3)=2\Delta n_g = n_{products(g)} - n_{reactants(g)} = 2 - (1 + 3) = -2. Given ΔH=92.38kJ=92380J\Delta H = -92.38 kJ = -92380 J, R=8.314JK1mol1R = 8.314 J K^{-1} mol^{-1}, T=298KT = 298 K. Using ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT: 92380=ΔU+(2×8.314×298)-92380 = \Delta U + (-2 \times 8.314 \times 298) 92380=ΔU4955.14-92380 = \Delta U - 4955.14 ΔU=92380+4955.14=87424.86J=87.42kJ\Delta U = -92380 + 4955.14 = -87424.86 J = -87.42 kJ

Explanation:

The change in internal energy is calculated by accounting for the work done by the contraction of gaseous moles during the synthesis of ammonia.

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