__Problem 1:__
The Critical constants for water are 647 K, 22.09 MPa and 0.0566 dm³mol⁻¹. Calculate the value of Van der Waals constants a and b.

We have Tc = 647 K , Pc = 22.09 MPa = 22.9 × 10³ KPa and Vc = 0.0566 dm³ mol⁻¹

__Answer:__We have Tc = 647 K , Pc = 22.09 MPa = 22.9 × 10³ KPa and Vc = 0.0566 dm³ mol⁻¹

Thus, b = Vc/3 = (0.0566 dm³ mol⁻¹)/3 =

**0.0189 dm³ mol⁻¹**
Again a = 3PcVc² = (22.9 × 10³ KPa) × (0.0566 dm³ mol⁻¹)² =

**212.3 KPa dm⁶ mol⁻¹**

__Problem 2:__
Calculate Van der Waals constants for Ethylene. (Tc = 280.8 K and Pc = 50 atm)

We know that, b = (1/8)(RTc/Pc)

Again, a = (27/64)(R²Tc²/Pc)

Argon has (Tc = - 122°C, Pc = 48 atm). What is the radius of the Argon atom?

We have Tc = - 122°C = (273 - 122) K = 151 K and Pc = 48 atm

We Know that, b = (1/8)(RTc/Pc)

∴ b = (1 × 0.082 lit atm mol⁻¹ K⁻¹ × 151 K)/(50 atm)

Again from Van der Walls equation, b = 4 × (4/3)πr³ × N₀

Thus, 4 × (4/3)πr³ × N₀ = (1 × 0.082 lit atm mol⁻¹ K⁻¹ × 151 K)/(50 atm)

__Answer:__We know that, b = (1/8)(RTc/Pc)

∴ b = (1 × 0.082 lit atm mol⁻¹ K⁻¹ × 280.8 K)/(8 × 50 atm)

**= 0.057 lit mol⁻¹**

Thus, a = {27 × (0.082 lit atm mol⁻¹ K⁻¹)² × (280.8 K)²}/(64 × 50 atm)

**= 4.47 lit² atm mol⁻²**

__Problem:__Argon has (Tc = - 122°C, Pc = 48 atm). What is the radius of the Argon atom?

__Answer:__We have Tc = - 122°C = (273 - 122) K = 151 K and Pc = 48 atm

We Know that, b = (1/8)(RTc/Pc)

∴ b = (1 × 0.082 lit atm mol⁻¹ K⁻¹ × 151 K)/(50 atm)

Again from Van der Walls equation, b = 4 × (4/3)πr³ × N₀

Thus, 4 × (4/3)πr³ × N₀ = (1 × 0.082 lit atm mol⁻¹ K⁻¹ × 151 K)/(50 atm)

**∴ r = 1.47 × 10⁻⁸ cm**