Question

A 1m^3 tank containing air at 10C and 350kPa is connectted through a valve to another tank containing 3 kg of air at 35C and 200kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20C. Determine the volume of the second tank an the final equilibrium pressure of air.

Answer 1

Given

Tank1

V1 = 1 m³

T1 = 10 C = 273 + 10 = 283 K

P1 = 350 Kpa = 350 * 10³ N/m²

R = 8.314 J/mol.K

PV = nRT

350 * 10³ N/m² * 1 m³ = n * 8.314 J/mol.K * 283 K

n1 = 148.75 moles

Tank 2

mass = 3 Kg = 3000 g

T2 = 35 C = 273 + 35 C = 308 K

P2 = 200 KPa = 200 * 10³ pa ( or N/m²

average molar mass of air = 28.97 g/mol

No. of moles n2 = mass / molar mass = 3000 g / 28.97 g/mol = 103.55 mole

P2 * V2 = n2 * R * T2

200 * 10³ N/m² * V2 = 103.55 mole * 8.314 J/mol.K * 308 K

V2 = 1.326 m³ Answer

equilibrium

when we add both the no. of moles n = n1 + n2 = 148.75 moles + 103.55 mole = 252.3 moles

total volume V= 1 + 1.326 = 2.236 m³

Given equilibrium temperature T = 20 C = 273 + 20 = 293 K

P * V = nRT

P * 2.236 m³ =  252.3 moles * 8.314 J/mol.K * 293 K

P = 274867.31 N/m² (pa) = 274.867 KPa Answer