**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 (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 (T
_{B}) 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 T _{B} 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=T _{B})**

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

**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=T _{B}=\( \frac{a}{Rb} \)**

**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.**

** **