Question

The van der Waals equation for 1 mole of gas is given by (p +av^-2)(v - b) = RT. In general, curves of p versus v forvarious values of T exhibit a maximum and a minimum at the twopoints where (δp/δv)_T = 0. The maximum and minumum coalesceinto a single point on that curve where(δ²p/δv²)_T = 0 inaddition to (δp/δv)_T = 0. This point is calledthe "critical point" of the substance and its temperature,pressure, and molar volume are denoted by T_c,p_c, and v_c, respectively.
    (a) Express a and b in terms of T_cand v_c.
    (b) Express p_c in terms ofT_c and v_c.
    (c) Write the van der Waals equation in terms ofthe reduced dimensionless variables T' ≡ T/T_c , v'≡ v/v_c , p' ≡ p/p_c
This form should involve neither a nor b.

Answer 1

Answer :

We have the van der walls equation

(P+a/V²)(V−b)=RT ................. (1)

After rearrangement we get

P = RT/(V−b) − a/V² ...................(2)

At critical Temperature, dP/dV = 0 and d²P/dV² = 0

Now starting from equation 2,

dP/dV = −RT/(V−b)² + 2a/V³ = 0 ....................(3)

or, a/V⁴ = RT/2V(V-b)² (rearranged) .................. (4)

Again starting from (3),

d²P/dV² = 2RT/(V-b)³ - 6a/V⁴ = 0 ................ (5)

Now putting the value from (4) we may rewrite

RT/(V-b)³ = 3RT/2V(V-b)²

or, V = 3b

We'll call it critical volume, V_c ; Hence V_c = 3b

Now we can put this value in equation (4)

a/81b⁴ = RT/2*3b*(3b-b)²

After simplifying we get T = 8a/27Rb, We call it critical temperature T_c.

Hence T_c = 8a/27Rb

Now if we put the values of P_c and T_c in equation (2),

we get critical pressure, P_c = a/27b²

Now we can express a and b in terms of V_c, T_c and P_c.

b = V_c /3

a = 27RbT_c/8 = 9RV_cT_c/8

again, a = 27b² P_c = 3 V_c² P_c (put b = V_c/3 )

Rewritting van der waals equation,

(P_c+a/V_c²)(V_c−b)=RT_c

or, (P_c+ 3 V_c² P_c /V_c²) (V_c− V_c /3) = RT_c

or, 4P_c * 2/3 V_c = RT_c

or, 8P_c V_c = 3RT_c

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