216.52L2atm/mol2 ??
Using van der waal's equation calculate the constant 'a' when two moles oaf a gas confined in a four litre flask exerts a pressure of a11.0 atm at a temperature of 300 K .
The value of b is 0.05 L/ mol.
-
UP 0 DOWN 0 0 8
8 Answers
106(11 + a*4/16)(4-2*0.05) = 2*0.0821*300
==> (11+a/4)*3.9 = 49.26
==> (11+a/4) = 12.63
==> a/4 = 1.63
==> a = 6.52 atm*L2mol-2
1Bhaiyya i had a doubt in this equation. Here why is it P+an^2/V^2 and not P-an^2/V^2 bcoz when the gas is real the intermolecular forces should decrease the pressure na, bcoz due to the forces the molecules wil collide with the walls with a less momentum n so the pressure shud b less. And this P is ideal pressure na?