$(a)\;0.99\;atm\qquad(b)\;17.85\;atm\qquad(c)\;34.96\;atm\qquad(d)\;18.20\;atm$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

$[P + \large\frac{a}{V^2}][V-b] = RT$

b is negligible

$\therefore PV + \large\frac{a}{V} = RT$

(or) $V^2P - RTV + a = 0$

$V = + RT \pm \large\frac{\sqrt{R^2T^2 - 4aP}}{2P}$

Since V has only one value

$R^2T^2 = 4aP$

$P = \large\frac{R^2T^2}{4a}$

$=\large\frac{{0.0821}^2\times273\times273}{4\times3.592}$

= 34.96 atm

Hence answer is (C)

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...