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
• 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 (0âž 100 atm),Â
⇒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
\(PV =Â RT (\frac{V}{V-b}) –Â \frac{a}{V} \)
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{\(T_B= \frac{a}{bR}\)}\)Â 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{\( PV=RT_B\)}\) Â , 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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