Question

Using van der waals equation calculate the constant 'a' when two moles of a gas confined in a four litre flask exerts a pressure of 11.0 atm at a temperature of 300K. The value of b is 0.05 lit mol^-1

Answer 1

$<b>HERE IS YOUR ANSWER$

----------------;------------------;--------------;--------------
----------------;------------------;--------------;--------------

➡$(P+an²/V²) (V-nb) = nRT$

$a = ( nRT-P / V - nb)V²/n² = ( 2mol × 0.08205L atm K^-1 mol^-1 × 300K - 11atm ÷ 4L - 2 mol × 0.05L mol^-1) × ( 4L)²÷ ( 2 mol)²$

➡ $6.49 atm L²mol^-2$




----------------;------------------;--------------;--------------
----------------;------------------;--------------;--------------

$<marquee>Hope it helps you$
$color{blue}{Thank\:you}$

Answer 2

Vander waal equation = [P + n²a/v²] [V - nb] = RT


According to question we have given;


P = 11 atm.

T = 300 K

V = 4 lit.

b = 0.05 lit. mol-¹

n = 2


Put the above values in Vander waal equation.

Then;


[11 + (2)² a/(4)²] [4 - (2 × 0.05)] = 2 × 0.082 × 300

[11 + 4a/16] [3.9] = 49.2

[11 + a/4] = 49.2/3.9

11 + a/4 = 12.61538

a/4 = 12.61538 - 11

a/4 = 1.61538

a = 1.61538 × 4


a = 6.46 atm. lit.² mol-²