Question

Twenty-five milliliters of liquid nitrogen (density = 0.807 g/mL) is poured into a cylindrical container with a radius of 11.3 cm and a length of 17.6 cm . The container initially contains only air at a pressure of 760.0 mmHg (atmospheric pressure) and a temperature of 309 K .

If the liquid nitrogen completely vaporizes, what is the total force (in lb) on the interior of the container at 309 K ?

Answer 1

Volume of the container = r²h

Where, r = radius of container and h is the length

r = 11.3 cm and h = 17.6 cm

So, Volume of the container (V) = (11.3 cm)²(17.6 . cm) = 7060.24 cm³ = 7060.24 mL

Volume of liquid nitrogen = 25 mL

Volume of container left free for N₂ = 7060.24 mL - 25 mL = 7035.24 mL = 7.035 L

Volume at STP is:

P₁V₁T₂ = P₂V₂T₁

Where, P₁,V₁ and T₁ are the pressure, volume and temperature of gas

But P₂,V₂ and T₂ are the pressure, volume and temperature of gas at STP

V₂ = ( P₁V₁T₁ / P₂ T₂) = (7.035 L * 760/760 atm * 273K)/ (1atm*309K) = 6.216 L

AT STP, 22.4 L of volume is occupied by 1 mol of N₂

6.216 L of volume is occupied by 1* 6.216L/22.4 L = 0.277 mol of N₂

Mass of N₂ = Volume * density = 25 mL * 0.807 g/mL = 20.175 g

Molar mass of N₂ = mass/Molar mass = 20.175g/ (28 g/mol ) = 0.720 mol

Total moles of gas = moles of N₂ +Moles of air = 0.720 + 0.277 = 0.9977 mol

FRom Ideal gas equation, PV = nRT

P = nRT/ V = ( 0.9977 mol* 8.314 J/mol.K * 309 K)/ 6.216 L = 412.36 J

Force( in lb) = 412.36 J *1lb/ 1.355 J = 304.32 lb force