3
For a batch process, the first law of thermodynamics states that, U =Q +W (1) Using equatio n 1 for U cal and since it is an adiabatic calorimeter (Q=0) and the wor e!erted by the stirrer is negligible (" =0), U cal =0 (#) $he energy balance in the bomb%s case is dE dt =d ( v 2 2 + gh+ U ) =dQ +dW (&)  'ssuming inetic and otential energy are relati*ely small dU =dQ +dW (+) ntegrating with "=0 since the bomb has a constant *olume (d-=0) , U B =Q B (.) For the U  of the entire process, U total =U system +U surroundings (/) which we can define the reactants (sample and o!ygen) to be the system and the rest of the calorimeter (bomb and water) to be the surroundings U total  is essentially ero gi*ing, U reactants = U calorimeter (2) where U surroundings  = 34*d$ from $1 to $# $he temperature change is relati*ely small and it is usually *alid to consider 4total to be constant 5ote that the negati*e sign indicates a decrease in energy of the process path

Bomb Calorimetry Experiment Derivation

Embed Size (px)

Citation preview

8/19/2019 Bomb Calorimetry Experiment Derivation

http://slidepdf.com/reader/full/bomb-calorimetry-experiment-derivation 1/3

For a batch process, the first law of thermodynamics states that,

∆ U =Q+W  (1)

Using equation 1 for ∆ U cal and since it is an adiabatic calorimeter (Q=0) and the wor

e!erted by the stirrer is negligible (" =0),

∆ U cal=0 (#)

$he energy balance in the bomb%s case is

dE

dt =d ( v

2

2+gh+U )=dQ+dW  (&)

 'ssuming inetic and otential energy are relati*ely small

dU =dQ+dW  (+)

ntegrating with "=0 since the bomb has a constant *olume (d-=0) ,

∆ U B=Q B (.)

For the ∆ U   of the entire process,

∆ U total=∆ U system+∆ U surroundings (/)

which we can define the reactants (sample and o!ygen) to be the system and the rest of the

calorimeter (bomb and water) to be the surroundings

∆ U total  is essentially ero gi*ing,

∆ U reactants=−∆ U calorimeter (2)

where ∆ U surroundings  = 34*d$ from $1 to $#

$he temperature change is relati*ely small and it is usually *alid to consider 4 total to be constant5ote that the negati*e sign indicates a decrease in energy of the process path

8/19/2019 Bomb Calorimetry Experiment Derivation

http://slidepdf.com/reader/full/bomb-calorimetry-experiment-derivation 2/3

6y definition,

dH =dU reactants+d ( PV ) (7)

8ifferentiating d(-),

d ( PV  )=VdP+ PdV    (9)

 'ssuming there is negligible e!pansion wor done,

dH =dU reactants+VdP (10)

$he products are then assumed to beha*e ideally to gi*e

d ( PV  )=V d nRT 

V  (11)

d ( PV  )= R d(nT ) (1#)

d ( PV  )= R ( ndT +Tdn ) (1&)

at constant temperature, d$=0,

d ( PV  )= RTdn

(1+)

:ubstituting equation 10 to 2 and integrating,

∫dH =∫dU +∫ RTdn (1.)

  ∆Ho = ∆Ureactants + RT∆n (16)

where ∆Ho is the standard enthalpy change, ∆U is the standard internal energy change, R is the

universal gas constant, T is the standard teperature and ∆n is the change in the nu!er o" 

oles during the process#

$o determine the 4bomb which is constant for both samples,

C bomb dT =C total dT −C water dT  (12)

8/19/2019 Bomb Calorimetry Experiment Derivation

http://slidepdf.com/reader/full/bomb-calorimetry-experiment-derivation 3/3

C bomb=∆ U B−mwater C water ∆ T 

∆ T  (17)

C bomb=Qba mba+5857.59(m0−m)

∆ T 

  −mwater C water (19)

where 4bomb  is the heat capacity of the calorimeter, Qba  is the heat liberated by the actual

combustion of benoic acid, mba is the mass of the sample, m 0 is the initial mass of the wire and

m is the mass of the wire after ignition, mwater  is the mass of water, 4water  is the heat capacity of 

water and ;$ is the change in temperature of water during the combustion process