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# FSC 1st Year Chemistry Chapter 7 Solved Exercise Notes PDF

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In this post”FSC 1st Year Chemistry Chapter 7 Solved Exercise Notes PDF” I am sharing PDF Notes for the students of FSC Part 1. The name of this Chapter is Thermodynamics . Students can download 1st Year Chemistry Chapter 7 Solved Exercise in PDF format of from here. This file contains 25 pages. These Chemistry Notes are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. Here are the complete 1st Year Chemistry Notes.

## 11th Class Chemistry Chapter 7 Thermochemistry Solved Exercise Notes PDF Download

### Explain that burning of a candle is a spontaneous process.

The process that makes place on its own is called spontaneous process. A process will also called spontaneous, if it needs energy to start with, but once it is started, then it proceed on its own.

A candle does not burn in air on its own; a spark initiates the burning of candle. Once it start burning, then the reaction proceed spontaneously to completion.

### What is the first law of thermodynamics? How does it explain that

• qv = DE                                                 (ii)               qp  =  DH

## First law of Thermodynamics:

It stated that:

“Energy can neither be created nor be destroyed but it can be changed from one form to another”

OR

“ Internal energy change of the system (DE) is equal to the sum of heat evolved or absorbed (q) and work sone by oron the system.

Mathematically’

DE     =   q + w

DE    =    q +PDV

1. qv = DE

Concider the heat supplied to the system at constant volume

DE    =     q +PDV   (at constant volume DV = 0 and PDV=0)

So,    DE    =  0

At constant volume Heat supplied is equal to internal energy change

1. qp  =  DH

For this, consider the enthalpy change at constant pressure. Enthalpy is the sum of internal energy and product of pressure and volume as

H = E. PV

Enthalpy change will be:

DH =DE+ D(PV)

DH =DE+PDV + DPV (since P= Constant DP=0 and DPV)

DH =DE+PDV

### For liquid and solid

DV=0 and PDV=0

So                           DH = DE

### For Gass; (since DV is not equal to 0)

DH = DE+ PDV

Put the value of   DE     =   q – w

When heat is supplied at constant pressure, work is done by the system

So                           w = – PDV

DE     =   q – PDV

DH = q – PDV + PDV

DH = qp

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