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Wize University Chemistry Textbook > Thermodynamics: Part 1: (0th law, 1st law, Calorimetry, Enthalpy)

Relating Calorimetry to the 1st Law

Relating Work and the Ideal Gas LawPrevious Section
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Calorimetry

  • We use calorimetry to measure the heat flow of a system. The following equation can be used to calculate heat flow using given that CC is the heat capacity of the calorimeter
q=CΔTq=C\Delta T
  • Heat capacity, CC, is dependent on the substance and is different depending on whether the calorimetry is performed at constant volume,CVC_Vor constant pressure,CPC_P
  • Specific heat capacity is expressed as a function of mass ( Cs or cC_s \ or \ c )
  • Molar heat capacity is expressed as a function of the number of moles ( CmC_m ).

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Case 1: Constant Volume Process, ΔV=0\Delta V=0

ΔV=0∴w=0∴ΔU=qvΔU=qv=∫T1T2nCV,mdT=nCV,m(Tf−Ti)\begin{array}{c} \Delta V=0\therefore w=0\therefore \Delta U=q_v\\[10pt] \displaystyle\Delta U=q_v=\int_{T_1}^{T_2}nC_{V,m}dT=nC_{V,m}(T_f-T_i) \end{array}

Note that we can use the exact same equations if we are told there is free expansion into a vacuum:
  • When expanding freely into a vacuum there is no external pressure
  • Since w=-PΔV and P=0, then w=0
pex=0∴w=0p_{ex}=0 \therefore w=0
ΔU=q+w∴ΔU=qv\Delta U=q+w\therefore\Delta U=q_v

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Case 2: Constant Pressure Calorimetry, Δp=0\Delta p=0

We need to define a new state function to keep track of heat exchange at constant pp
H=U+pVΔH=qp=∫T1T2nCp.mdT=nCp,m(Tf−Ti)\begin{array}{c} H=U+pV\\[10pt] \Delta H=q_p=\displaystyle\int_{T_1}^{T_2}nC_{p.m}dT=nC_{p,m}(T_f-T_i) \end{array}

It can be shown that
Cp,m=CV,m+RC_{p,m}=C_{V,m}+R


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Case 3: Isothermal Conditions

Under isothermal conditions (constant temperature), ∆T=0

Wize Tip
∆U in the equation: ∆U= q + w, depends only on temperature!
So when there is no temperature change under isothermal conditions, there is no change in U ( ∆U=0)

Note: because it’s isothermal, ∆U=0
ΔU=q+w=0∴q=−w\Delta U=q+w=0 \therefore q=-w

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Heat Capacity of Ideal Gases

  • It can be shown that
Cp,m=CV,m+RC_{p,m}=C_{V,m}+R
  • For an ideal monoatomic gas
CV,m=32RCp.m=52RC_{V,m}=\dfrac{3}{2}R\quad C_{p.m}=\dfrac{5}{2}R
  • For an ideal diatomic gas
CV,m=52RCp.m=72RC_{V,m}=\frac{5}{2}R\quad C_{p.m}=\frac{7}{2}R

The following is low yield info that can help you memorize these equations (but they will most likely be provided to you on a formula sheet).
  • Monoatomic gases have 3 degrees of freedom (they are able to move in 3 directions)
  • While diatomic gases have 5 degrees of freedom (they are able to move in 5 directions)
  • The degrees of freedom matches the numerator of the Cv equation for each!

The enthalpy of combustion for ethanol C2H5OH is -1370.7 kJ/mol. Calculate q and ΔH when 45 g of ethanol are burned at 1 atm and 298 K.
  1. q (kJ)
  2. deltaH (kJ)
(Don't include units in your answer and round to the nearest whole number)
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Practice: Isothermal Expansion

4 moles of neon was confined to a 8L flask initially at room temperature underwent an isothermal expansion into a vacuum at 348K. Calculate ∆𝑈, ∆𝐻, 𝑎𝑛𝑑 𝑞 for this process.