Using van der Waal's equation of state, calculate the pressure correction factor (in atm L 2 mol -2 ) for two moles of a gas confined in a 4-litre flask that exert a pressure of 11 atm at 300 K , b=0.5 L mol -1 Use: R=0.08 litre atm mol -1 K -1

Question

Using van der Waal's equation of state, calculate the pressure correction factor (in atm L 2 mol -2 ) for two moles of a gas confined in a 4-litre flask that exert a pressure of 11 atm at 300 K , b=0.5 L mol -1 Use: R=0.08 litre atm mol -1 K -1 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(P+(n2 a/V2))(V-n b) =n R T ⇒ (n2 a/V2)=(n R T/V-n b)-P =(2 × 0.08 × 300/4-2 × 0.5)-11 =12-11=1 atm L 2 mol -2

Q. Using van der Waal's equation of state, calculate the pressure correction factor (in $a t m L^{2} m o l^{- 2}$ ) for two moles of a gas confined in a $4$ -litre flask that exert a pressure of $11 a t m$ at $300 K, b = 0.5 L m o l^{- 1}$ Use: $R = 0.08$ litre atm $m o l^{- 1} K^{- 1}$

$(P + \frac{n ^{2} a}{V ^{2}}) (V - nb) = n RT$
$\Rightarrow \frac{n ^{2} a}{V ^{2}} = \frac{n RT}{V - nb} - P = \frac{2 \times 0.08 \times 300}{4 - 2 \times 0.5} - 11$
$= 12 - 11 = 1 a t m L^{2} m o l^{- 2}$

Q. Using van der Waal's equation of state, calculate the pressure correction factor (in atmL2mol−2 ) for two moles of a gas confined in a 4-litre flask that exert a pressure of 11atm at 300K,b=0.5Lmol−1 Use: R=0.08 litre atm mol−1K−1

(P+V2n2a​)(V−nb)=nRT ⇒V2n2a​=V−nbnRT​−P=4−2×0.52×0.08×300​−11 =12−11=1atmL2mol−2

Q. Using van der Waal's equation of state, calculate the pressure correction factor (in $a t m L^{2} m o l^{- 2}$ ) for two moles of a gas confined in a $4$ -litre flask that exert a pressure of $11 a t m$ at $300 K, b = 0.5 L m o l^{- 1}$ Use: $R = 0.08$ litre atm $m o l^{- 1} K^{- 1}$

$(P + \frac{n ^{2} a}{V ^{2}}) (V - nb) = n RT$
$\Rightarrow \frac{n ^{2} a}{V ^{2}} = \frac{n RT}{V - nb} - P = \frac{2 \times 0.08 \times 300}{4 - 2 \times 0.5} - 11$
$= 12 - 11 = 1 a t m L^{2} m o l^{- 2}$