Application of Vanderwaals Equation in Calculation of Boyle’s temperature

Application of Vanderwaals Equation in Calculation of Boyle’s temperature

Consider the Z vs P plots of nitrogen at different temperatures varying between – 70°C and 50 °C

Boyle's temperature
• The graph shows that as the temperature is raised, the dip in the curve becomes smaller and smaller.

At 50°C the curve is almost horizontal for an appreciable range of pressure (0➠100 atm), 
⇒Z ≈ 1 under these conditions.
In other words, PV = constant and thus Boyle’s law is obeyed within this range of pressure at 50°C.
• This temperature is called the Boyle point or Boyle temperature.
• Boyle’s temperature is that temperature at which a real gas obeys the ideal gas law over an appreciable range of pressure.
• Below Boyle’s temperature, the value of Z is firstly decreases, approaches a minimum and then increases continuously with increase in the pressure.

• Above Boyle’s temperature (above 50°C) ,the value of compressibility factor (z) shows a continuous rise with increase in pressure.

  • As shown in the fig.,from the curve of z(\( \frac{PV}{nRT} \)) vs P, the minima lie on a parabolic curve.
  • On increasing the temperature, dip in the curve decreases and at a certain temperature the dip disappears.
  • This temperature is called as Boyle’s temperature (TB) and at this temperature the real gas obeys ideal gas law.

At an intermediate temperature, the value of z \(\approx\) 1 over a large range of pressure

i.e., at Boyle’s temperature, a real gas obeys Boyle’s law (PV=constant) for a large range of pressure.

Mathematically Boyle’s temperature is defined as the temperature at which

• \(\lim_{P\to0}[\frac{∂(PV)}{∂P}=0\)

The Boyle’s temperature TB can easily be be obtained from VWL equation:
As we know, the van der Walls equation for 1 mole of a gas is,
\( \left( {P + \frac{{a}}{{V^2 }}} \right)\left( {V – b} \right) = RT \)

⇒ \(P =  \frac{RT}{V-b} –  \frac{a}{V^2} \)

Multiplying both sides by V, we get
\(PV =  RT (\frac{V}{V-b}) –  \frac{a}{V} \)

Differentiating above equation w.r.t. P at constant temperature,

\(\frac{∂(PV)}{∂P}=RT[ (\frac{1}{V-b}) (\frac{∂V}{∂P})_T+\frac{V(-1)}{(V-b)^2} (\frac{∂V}{∂P})_T] + \frac{a}{V^2}(\frac{∂V}{∂P})_T\)

\(=RT[ \frac{1}{(V-b)}-\frac{V}{(V-b)^2}](\frac{∂V}{∂P})_T + \frac{a}{V^2}(\frac{∂V}{∂P})_T\)

\(=[ \frac{RT}{(V-b)}-\frac{RTV}{(V-b)^2} + \frac{a}{V^2}](\frac{∂V}{∂P})_T\)

For minima, put \(\frac{∂(PV)}{∂P}=0\)

\(\frac{RT}{(V-b)}-\frac{RTV}{(V-b)^2} + \frac{a}{V^2}=0\)

\(RT[\frac{1}{(V-b)}-\frac{V}{(V-b)^2}] =-\frac{a}{V^2}\)

\(RT[\frac{-b}{(V-b)^2}] =-\frac{a}{V^2}\)

\(RT= \frac{a}{b} (\frac{V-b}{V})^2\)

\( T=\frac{a}{bR}(1-\frac{b}{v})^2 \)

At Boyle’s temperature(T=TB)
\( T_B=\frac{a}{bR}(1-\frac{b}{v})^2 \)

When \(P \to 0\) , the volume \(V\to\infty\)
\(\therefore \frac{b}{V}=0\) & our equation reduces to,

\(T_B= \frac{a}{bR}\)

\( \fbox{\(T_B= \frac{a}{bR}\)}\) viz.,the Boyle’s temp.

To Prove : At Boyle’s temperature, the VWL equation is reduced to Ideal gas equation:

Proof:
As we know, VWL equation for one mole of a gas is,
\( \left( {P + \frac{{a}}{{V^2 }}} \right)\left( {V – b} \right) = RT \)

⇒\(PV= RT- \frac{a}{V}+bP+\frac{ab}{V^2}\)

At moderate pressures, V>>>a & b ∴ we can neglect the term \( \frac{ab}{V^2} \)

⇒\(PV=RT- \frac{a}{V}+bP \)

Replacing V by \( \frac{RT}{P} \) in term \( \frac{a}{V} \), we have

\( PV=RT-\frac{aP}{RT}+bP \)

⇒\( PV=RT+P(b-\frac{a}{RT})  \tag{1}\\ \)

At Boyle’s temperature,T=TB=\( \frac{a}{Rb} \)
Hence equation 1 becomes ,

\(\fbox{\( PV=RT_B\)}\)  , viz., the ideal gas equation.

Hence at Boyle’s temperature, the van der Walls equation is reduced to ideal gas equation.

 

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