Deduce the value of critical constant of van der waals equation
Answers
Answer:
The critical temperature, Tc, is characteristic of every gas and may be defined as: “The temperature below which the continuous increase of pressure on a gas ultimately brings about liquefaction and above which no liquefaction can take place no matter what so ever pressure be applied”.
The pressure required to liquefy the gas at critical temperature is called critical pressure and the volume occupied by 1 mole of gas under these conditions is called the critical volume.
Van der Waals equation may be written as:
\left(P + \dfrac{a}{V^2} \right) (V- b) = RT \\[3mm] \text{or} \hspace{20mm} PV + \dfrac{a}{V}- pb- \dfrac{ab}{V^2} = RT \\[3mm] \text{or} \hspace{10mm} PV^3 + Av- PbV^2- ab- RTV^2 = 0 \\[3mm] \text{or} \hspace{10mm} V^3- \left( \dfrac{RT}{P} + b \right)V^2 + \dfrac{Av}{P}- \dfrac{ab}{P} = 0 ……..(iii)
This is the general gas equation of van der Waals. Since this equation is cubic hence it should have three values of volume, V, for a given pressure. At critical temperature all the three values of V become identical. This is due to the fact that at critical temperature and pressure the distinction between liquid and vapour disappears. The identical volume, V is called critical volume, Vc.
i.e.
V = V_c
Or V- V_c = 0
Or (V- V_c)^3 = 0
Or V^3- 3V^2V_c + 3VV_c^2- V_c^3 = 0………….(iv)
This is called an equation of critical state. Substituting T = T_c (critical temperature) and P = P_c (critical pressure) in equation (iii), we get,
V^3- V^2 \left(\dfrac{RT_c}{P_c} + b \right) + V. \dfrac{a}{P_c}- \dfrac{ab}{P_c} = 0 …………(v)
Comparing eqns. (iv) and (v) we get
3V_c = \dfrac{RT_c}{P_c} + b …………(vi)
3V^2_c = \dfrac{a}{P_c}………….(vii)
V^3c = \dfrac{ab}{P_c} …………….(viii)
On solving the above equations, we get
V_c = 3b \cdots (ix) \\[3mm] P_c = \dfrac{a}{27b^2} \cdots (x) \\[3mm] T_c = \dfrac{8a}{27b R} \cdots (xi)
Since it is difficult to find out the accurate value of V_c ,it is better to express ‘a’ and in terms of P_c and T_c. Substituting the value of from eq. (ix) in \text{(vi), we get } 3V_c = \dfrac{RT_c}{P_c} + \dfrac{V_c}{3} \\[3mm] \text{or} \hspace{20mm} \dfrac{8}{3}V_c = \dfrac{RT_c}{P_c} \\[3mm] \text{or} \hspace{20mm} V_c^2 = \dfrac{9}{64}. \dfrac{R^2T^2_c}{P_c^2} \cdots (xii)
Now, substituting the value of V^2_c from eq. (xii) in eq. (vii),
\dfrac{a}{2P_c} = \dfrac{9}{64}. \dfrac{R^2T^2_c}{P_c} \\ \text{or} \hspace{20mm} a = \dfrac{27}{64}. \dfrac{R^2T^2_c}{P_c} \cdots (xii)
On substituting the value of V_c from eq. (ix) in eq. (vi), we get
3 \times 3b = \dfrac{RT_c}{P_c} + b \\ b = \dfrac{RT_c}{8P_c} ……………….(xiv)