Chapter 6, Problem 6.28QP

Interpretation Introduction

Interpretation:

The work done (in Joules) and $\text{ΔU}$ against given pressure and given temperature has to be calculated.

Concept Introduction:

The work done on the system or by the system can be calculated from the equation

$\text{w=-PΔV}$

Where w=work done

P=Pressure

V=Volume

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$\text{H=U+PV}$

Where, H=enthalpy

U=internal energy

P=Pressure

V=Volume

For any process, the enthalpy change is given as,

$\text{ΔH=ΔU+Δ(PV)}$

If the pressure is held constant,

$\text{ΔH=ΔU+PΔV}$

Where,$\text{ΔH}$= change in enthalpy

$\text{ΔU}$ =change in internal energy

P=Pressure

$\text{ΔV}$ = change in volume