Question

Using van der waal equation , calculate the constant a when two moles of gas confined in a 4 L flash exert a pressure of 11 atm at temperature of 300K. The value of b is 0.05 L mol^-1 .

Answer 1

Answer:

$\display \text{a = 6.46 atm L^2 \ mol^{-2} }$

Explanation:

Given :

No. of moles n = 2 moles

Volume V = 4 L

Pressure P = 11 atm

Temperature T = 300 K

Value of b = 0.05 L mol⁻¹

We have van der waal equation

$\displaystyle{\left(P-\dfrac{an^2}{V^2}\right)(V-nb)=nRT}$

Putting value here :

$\displaystyle{\left(11+\dfrac{a\times2^2}{4^2}\right)(4-2\times0.05)=2\times0.082\times300}\\\\\\\displaystyle{\left(11+\dfrac{4a}{16}\right)(4-0.1)= 0.082\times600}\\\\\\\displaystyle{\left(\dfrac{176+4a}{16}\right)(3.9)=49.2}\\\\\\\displaystyle{15.6a=787.2-686.4}\\\\\\\display \text{a = 6.46 atm L^2 \ mol^{-2} }$

Hence we get answer .

Answer 2

van der waals equation for n moles of gas is

[P + n2a/V2] [V – nb] = nRT

Given V = 4 litre; P = 11.0 atm, T = 300 K;

b = 0.05 litre mol-1, n = 2 ;

thus, [11 + 22a/42] [4 – 2 *0.05] = 2 * 0.082 *300

∴ a = 6.46 atm litre2 m