Chemistry Daily Challenge 25-July-2015

Chemistry Level 2

van der Waals equation is ( p + a V 2 ) ( V − b ) = R T \left(p+\frac{a}{V^2}\right)\left(V-b\right)=RT Given: ∂ P ∂ V = 0 a n d ∂ 2 P ∂ V 2 = 0 at critical point \frac{\partial P}{\partial V}=0 \quad and \quad \frac{\partial^2 P}{\partial V^2}=0 \quad \text{at critical point}

If a and b can be expressed as a = p R 2 T c 2 q P c b = r R T c s P c a=\frac{pR^2T_c^2}{qP_c}\quad b=\frac{rRT_c}{sP_c}

where T c , P c , V c T_c,P_c,V_c are critical properties. p and q coprime integer.r and s are coprime integer. Find the value of p+q+r+s. \color{#3D99F6}{\text{Find the value of p+q+r+s.}}


The answer is 100.

Aamir Faisal Ansari by Aamir Faisal Ansari , 24, India

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1 solution

Using Van Der Waal's Equation,

P = R T V − b \frac{RT}{V-b} - a V 2 \frac{a}{V^{2}}

Taking the partial derivative and equating to zero,

R T ( V − b ) 2 \frac{RT}{(V-b)^{2}} = 2 a V 3 \frac{2a}{V^{3}} ...... (1)

Taking the partial derivative again and equating it to zero,

2 R T ( V − b ) 3 \frac{2RT}{(V-b)^{3}} = 6 a V 4 \frac{6a}{V^{4}} ..... (2)

Dividing the 2 equations,

We will get the result Critical Volume is 3b.

Using the above result in (1), we get Critical Temperature is 8 a 27 R b \frac{8a}{27Rb}

Using both these results in Van der waals equation, we get Critical Pressure is a 27 b 2 \frac{a}{27b^{2}}

Using these results in the parameters given in the question, it follows that:

a = p R 2 64 a 2 27 b 2 729 R 2 b 2 \frac{pR^{2}64a^{2}27b^{2}}{729R^{2}b^{2}} and b = r R 8 a 27 b 2 s 27 R b a \frac{rR8a27b^{2}}{s27Rba}

p q \frac{p}{q} = 64 27 \frac{64}{27} and r s \frac{r}{s} = 8 1 \frac{8}{1}

Since p,q and r,s are co prime integers, p = 64 q = 27 r = 8 s = 1

p + q + r + s = 64 + 27 + 1 + 8 = 100 \boxed{100}

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Moderator note:

Simple standard approach.

Good solution.This isn't a chemistry prob.Already gave the first and second derivatives are zero.

Spandan Senapati - 4 years, 3 months ago

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