Question

Steam at 275ºC and 15.0 bar is contained in a rigid tank with the volume of 10.0 m^3 . In a constant volume cooling process, the pressure inside the tank drops to 1.8 bar and some of the steam condenses. Calculate 1) the temperature of the tank contents after the process, 2) the amount of steam condensed during the process (in kg), and 3) the amount of heat removed from the tank during the process.

Answer 1

Given initial temperature, T₁ = 275 ^oC = 275 +273 = 548 K

initial pressure P₁ = 15 bar

initial volume V₁ = 10 m³

final temperature, T₂ = ?

final pressure P₂ = 1.8 bar

final volume V₂ = 10 m³

Applying combined gas equation P₁V₁/T₁ = P₂V₂/T₂

or T₂ = P₂V₂ T₁/ (P₁V₁) = 1.8 X 10 X 548 /. (15 X 10) = 65.78 K

Let initial number of moles of gaseous moles is n₁

then P₁V₁ = n₁RT₁

or n₁ = P₁V₁/RT₁ = 15 X 10 / (8.314 × 10⁻⁵ X 548) = 3293.3 moles

(value of R = 8.314 × 10⁻⁵ m³ bar K⁻¹ mol⁻¹ )

Let initial number of moles of gaseous moles is n₂

then P₂V₂ = n₂RT₂

or n₂= P₂V₂/RT₂ = 1.8 X 10 / (8.314 × 10⁻⁵ X 65.78) = 3291.3 moles

Then number of moles of steam condensed = n₁-n₂ = 3293.3 moles - 3291.3 moles = 2 moles

Molar mass of steam = 18 g/mol

Hence mass of steam condenced = 18 g/mol X 2 mol = 36 g = 0.036 kg