The pressure exerted by 1 mol of CO 2 at 273 K is 34.98 atm. Assuming that volume occupied by CO 2 molecules is negligible, the value of van der Waals' constant for attraction of CO 2 gas is

Question

The pressure exerted by 1 mol of CO 2 at 273 K is 34.98 atm. Assuming that volume occupied by CO 2 molecules is negligible, the value of van der Waals' constant for attraction of CO 2 gas is (A) 3.59 dm 6 atm mol -2 (B) 2.59 dm 6 atm mol -2 (C) 1.25 dm 6 atm mol -2 (D) 1.59 dm 6 atm mol -2

Tardigrade Admin · Accepted Answer

[P+(a/V2)][V-b]=R T ∴[P+(a/V2)] V=R T Or V2 P-R T V+a=0 V=(+R T ± √R2 T2-4 Pa /2 P) Since, V is constant at given P and T, V can have only one value or discriminant =0 ∴ R2 T2=4 Paor a=(R2 T2/4 P) =((0.821)2 ×(273)2/4 × 34.98) =3.59 dm 6 atm mol -2

Q. The pressure exerted by $1 m o l$ of $C O_{2}$ at $273 K$ is $34.98 a t m$ . Assuming that volume occupied by $C O_{2}$ molecules is negligible, the value of van der Waals' constant for attraction of $C O_{2}$ gas is

$3.59 d m^{6} a t m m o l^{- 2}$

$2.59 d m^{6} a t m m o l^{- 2}$

$1.25 d m^{6} a t m m o l^{- 2}$

$1.59 d m^{6} a t m m o l^{- 2}$

Q. The pressure exerted by 1mol of CO2​ at 273K is 34.98atm. Assuming that volume occupied by CO2​ molecules is negligible, the value of van der Waals' constant for attraction of CO2​ gas is

3.59dm6atmmol−2

2.59dm6atmmol−2

1.25dm6atmmol−2

1.59dm6atmmol−2

[P+V2a​][V−b]=RT ∴[P+V2a​]V=RT Or V2P−RTV+a=0 V=2P+RT±R2T2−4Pa​​ Since, V is constant at given P and T,V can have only one value or discriminant =0 ∴R2T2=4 Paor a=4PR2T2​ =4×34.98(0.821)2×(273)2​ =3.59dm6atmmol−2

Q. The pressure exerted by $1 m o l$ of $C O_{2}$ at $273 K$ is $34.98 a t m$ . Assuming that volume occupied by $C O_{2}$ molecules is negligible, the value of van der Waals' constant for attraction of $C O_{2}$ gas is

$3.59 d m^{6} a t m m o l^{- 2}$

$2.59 d m^{6} a t m m o l^{- 2}$

$1.25 d m^{6} a t m m o l^{- 2}$

$1.59 d m^{6} a t m m o l^{- 2}$