Question

Using Van der Waals equations, calculate the constant $a$ when 2 moles of a gas confined in a 4$-litre$ flask exerts a pressure of 11.0 atm at a temperature of 300 K. The value of $b$ is 0.05 $litremo{l}^{-1}$.

• A. $29.41c{m}^3$
• B. $58.82c{m}^3$
• C. $117.64c{m}^3$
• D. $6.52c{m}^3$

Answer 1

The correct option is D $6.52c{m}^3$

$(P+\frac{a{n}^2}{V^2})(V-nb)=nRT$

$(11+\frac{a(2{)}^2}{(4{)}^2})(4-2\times 0.05)=2\times 0.0821\times 300$

$(11+\frac{a}{4})(4-0.1)=49.26$

$a=6.523c{m}^3$

Hence, the correct answer is option $\text{D}$.