Thermodynamics

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Payal Gupta

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1. (ii) whose value is independent of path

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alok kumar singh

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This is a Short Answer Type Questions as classified in NCERT Exemplar

At constant volume

q = ΔU + (-w)

-w = pΔ q = AU + pΔV

ΔV = 0 (at constant volume)

Hence, qv = ΔU + 0 = ΔU= change in internal energy at constant pressure, qp = AU + pΔV

Since ΔU + pΔV=ΔH

=> qp = ΔH change in enthalpy

Hence, at constant volume and at constant pressure, heat change is a state function because it is equal to ΔU and ΔH respectively which are state functions.

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alok kumar singh

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For an isolated system w = 0, q = 0

Since ΔU= q + w = 0 + 0 = 0, ΔU= 0

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alok kumar singh

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Gibbs energy for a reaction in which all reactants and products are in standard state. ΔrG° is related to the equilibrium constant of the reaction as follows

ΔrG = ArG° + RT In K

At equilibrium, 0 = ΔrG° + RT InA– ( {ΔrG = 0) or  ΔrG° =-RT lnK

ΔrG° = 0 when K= 1

For all other values of K, ArG° will be non-zero.

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alok kumar singh

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Molar enthalpy of vaporization is more for water due to hydrogen bonding between water molecules.

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alok kumar singh

Contributor-Level 10

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State functions: Enthalpy, entropy, temperature, free energy Path functions: Heat, work.

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alok kumar singh

Contributor-Level 10

This is a Short Answer Type Questions as classified in NCERT Exemplar

The standard molar entropy of H20 (1) is 70 J K-1 mol-1. The solid form of H20 is ice. In ice, molecules of H20 are less random than in liquid water. Thus, molar entropy of H20 (s) < molar entropy of H20 (1). The standard molar entropy of H20 (s) is less than 70 J K-1 mol-1.

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