One mole of a van der Waal's gas at 27° C and 960 atm occupies 0.075 L. If b=0.025 L mol -1 then value of compressibility factor is

Question

One mole of a van der Waal's gas at 27° C and 960 atm occupies 0.075 L. If b=0.025 L mol -1 then value of compressibility factor is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

P text ideal (V-b)=R T ⇒ P text ideal =(0.08 × 300/0.075-0.025) =(24/0.05)=480 atm ⇒ Z =[( P text real / P text ideal )] T , V =(960/480)=2

Q. One mole of a van der Waal's gas at $2 7^{\circ} C$ and $960 a t m$ occupies $0.075 L$ . If $b = 0.025 L m o l^{- 1}$ then value of compressibility factor is

$P_{ideal} (V - b) = RT$
$\Rightarrow P_{ideal} = \frac{0.08 \times 300}{0.075 - 0.025}$
$= \frac{24}{0.05} = 480 a t m$
$\Rightarrow Z = [\frac{P _{real}}{P _{ideal}}]_{T, V} = \frac{960}{480} = 2$

Q. One mole of a van der Waal's gas at 27∘C and 960atm occupies 0.075L. If b=0.025Lmol−1 then value of compressibility factor is

Pideal ​(V−b)=RT ⇒Pideal ​=0.075−0.0250.08×300​ =0.0524​=480atm ⇒Z=[Pideal ​Preal ​​]T,V​=480960​=2

Q. One mole of a van der Waal's gas at $2 7^{\circ} C$ and $960 a t m$ occupies $0.075 L$ . If $b = 0.025 L m o l^{- 1}$ then value of compressibility factor is

$P_{ideal} (V - b) = RT$
$\Rightarrow P_{ideal} = \frac{0.08 \times 300}{0.075 - 0.025}$
$= \frac{24}{0.05} = 480 a t m$
$\Rightarrow Z = [\frac{P _{real}}{P _{ideal}}]_{T, V} = \frac{960}{480} = 2$