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Thermodynamic Processes

1. Isochoric Processes (constant volume ΔV=0\Delta V=0)
  • For any point during the process PT=Constant\frac{P}{T}=Constant dW=Pdv=0W=0dW=Pdv=0\to W=0 ΔU=32nRΔT=32NkBΔT\Delta U=\frac{3}{2}nR\Delta T=\frac{3}{2}Nk_B\Delta T for monatomic gases ΔU=QWΔU=Q=nCvΔT\Delta U=Q-W\to\Delta U=Q=nC_v\Delta T where CvC_v is molar specific heat at constant volume
  • CvC_v is equal to 32R\frac{3}{2} R, 52R\frac{5}{2} R and 72R\frac{7}{2} R for monatomic, diatomic and multi-atomic gases respectively.
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2. Isobaric Processes (Constant pressure ΔP=0\Delta P=0)
  • For any point during the process VT=Constant\frac{V}{T}=Constant W=dW=PdV=PΔV=nRΔTW=\int_{ }^{ }dW=\int_{ }^{ }PdV=P\Delta V=nR\Delta T Q=nCpΔTQ=nC_p \Delta T where CpC_p is molar specific heat at Constant Pressure and is equal to Cv+RC_v+R
  • The change in the internal energy is always equal to nCvΔTnC_v \Delta T independent of the type of the process.
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3. Isothermal processes (Constant Temperature ΔT=0\Delta T=0)
  • For any point during this process PV=ConstantPV=Constant ΔU=nCvΔT=0Q=WW=pdV=nRTVdV=nRTViVfdVV=nRTln(VfVi)\begin{aligned} & \Delta U=nC_v \Delta T=0\to Q=W\\ & W=\int pdV=\int \dfrac{nRT}{V} dV=nRT\int_{V_i}^{V_f} \dfrac{dV}{V}=nRT \ln⁡\Big(\dfrac{V_f}{V_i}\Big) \end{aligned}
  • If compression or expansion happens very slowly, the gas can exchange heat with the environment to keep its temperature constant.
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4. Adiabatic Processes (no heat transfer Q=0Q=0)
  • Q=0ΔU=W=nCVΔTQ=0\to\Delta U=-W=-nC_V \Delta T For any point during an adiabatic process PVγ=ConstantPV^\gamma=Constant In addition, for any point TVγ1=ConstantTV^{\gamma-1}=Constant' where γ=CpCv\gamma=\frac{C_p}{C_v} (γ=53\gamma=\frac{5}{3} for monatomic gases )
  • If compression or expansion happens quickly, the gas does not have time to transfer heat to the environment so, Q=0Q=0 and the process is adiabatic.
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Key Terms

The following lesson is low yield in terms of exams. We'll just focus on understanding. 💡

System Vs Surroundings

System-reactants and products that are part of the chemical reaction







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Open Vs Closed Vs Isolated Systems


Examples:
  • open system → cup of coffee
  • closed system → bottle of pop
  • isolated system → perfect thermos
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Example: Adiabatic Processes

The pressure of diatomic ideal gas triples due to an adiabatic process. If the final temperature of the gas is 320 K, what was the initial temperature?

pf=3 pip_f=3 \ p_i
Tf=320KT_f=320K
Ti=?T_i=?

We know that:
piViγ=pfVfγ               &                 TiViγ1=TfVfγ1p_iV_i^{\gamma}=p_fV_f^{\gamma} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_iV_i^{\gamma-1}=T_fV_f^{\gamma-1}
Isolate what we are solving for
Ti=Vfγ1Viγ1Tf=(VfVi)γ1TfT_i=\cfrac{V_f^{\gamma-1}}{V_i^{\gamma-1}}T_f=\big(\frac{V_f}{V_i}\big)^{\gamma -1}T_f
NeedVfVi=?   WellVfγViγ=pipfNeed\cfrac{V_f}{V_i}=?\ \ \ Well\cfrac{V_f^{\gamma}}{V_i^{\gamma}}=\cfrac{p_i}{p_f}
   (VfVi)γ=pipf\to\ \ \ \big(\frac{V_f}{V_i}\big)^{\gamma }=\frac{p_i}{p_f}
Need to use exponent rules to get rid of 𝛾
[(VfVi)γ]1/γ    (VfVi)γ/γ=(pipf)1/γ    remember that  γγ=1\bigg[\big(\frac{V_f}{V_i}\big)^{\gamma}\bigg]^{1/\gamma}\ \ \ \ \to \bigg(\frac{V_f}{V_i}\bigg)^{\gamma/ \gamma}=\bigg(\frac{p_i}{p_f}\bigg)^{1/ \gamma}\ \ \ \ remember\ that\ \ \frac{\gamma}{\gamma}=1
VfVi=(pipf)1/γ\frac{V_f}{V_i}=\big(\frac{p_i}{p_f}\big)^{1/\gamma} Good! Now we may find T;
Ti=[(pipf)1/γ]γ1    Tf=(pipf)γ1/γ×TfT_i=\bigg[\big(\frac{p_i}{p_f}\big)^{1/\gamma}\bigg]^{\gamma-1}\ \ \ \ T_f=\big(\frac{p_i}{p_f}\big)^{\gamma-1/\gamma}\times T_f
𝛾= 1.4 for diatomic ideal gas
Ti=(13)1.41/1.4×320K=233.8KT_i=\big(\frac{1}{3}\big)^{1.4-1/1.4}\times 320K=233.8K

A system composed of 4.50 L of N2 in a cylinder expands against an external pressure of 300 kPa until its volume is 6.30 L.
  1. What is the value of w for this process?
  2. What would the value of w be if the external pressure was 0 kPa?
Extra Practice