Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Using Van der Waals's equation . Calculate the constant 'a' when two mole of a gas confined in a four litre flask exerts a pressure of 11.0 atmospheres at a temperature of 300K . The value of b is 0.05 lit $mol^{-1}$

$\begin{array}{1,1}(a)\;646 atm\;litre^2\;mol^{-2}\\(b)\;6.46\;atm\;litre^2\;mol^{-2}\\(c)\;64.6 atm\;litre^2\;mol^{-2}\\(d)\;0.64atm\;litre^2\;mol^{-2}\end {array}$

Can you answer this question?

1 Answer

0 votes
Van der Waal's equation for n mole of gas is
[P + $\large\frac{n^2a}{V^2}$] [ V-nb] = nRT
V = 4 litre
P = 11.0 atm
T = 300K
b = 0.05 litre$\;mol^{-1}$
n = 2
[ 11+ $\large\frac{2^2a}{4^2}][4-2\times0.05]$
$a = 6.46 \;atm\;litre^2mol^{-1}$
Hence answer is (B)
answered Apr 2, 2014 by sharmaaparna1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App