1. Pressurized tanks are commonly used in gas propulsion for space applications.

Question

1.      Pressurized tanks are commonly used in gas propulsion for space applications. A closed, rigid and insulated tank with a volume of 0.1 m3contains 0.7 kg of N2 and 1.1 kg of CO2at 27oC. Determine the following:

a.       The composition in terms of mass fractions

b.      The composition in terms of mole fractions

c.       The partial pressure of each component

d.      The mixture pressure

e.       The heat transfer required to bring the mixture to 127oC

f.       The entropy change of the mixture

Explanation / Answer

a) Mass fraction (Wi) can be calculated by the equation Wi= mi/M, where mi is the mass of 'i' th substance, M total mass of gas mixture.

So, Wnitrogen =700g/ (700g+1100g) = 0.39

WCO2 = 1100g/ (700g+1100g) = 0.61

b) Mole fraction (Xi) can be calculated by the equation Xi= ni/N, where ni is the mole number of 'i' th substance, N total mole number of gas mixture

Again mole number of a gas in a gas mixture ni= mass in g/ molar mass in g

So, nnitrogen =700g/ 28g = 25 mole; molar mass of nitrogen is 28 g

nCO2 = 1100g/ 44g = 25 mole; molar mass of carbondioxide is 44 g

so, Xnitrogen= 25/(25+25)= 0.5

XCO2= 25/(25+25)= 0.5

d) Mixture pressure can be calculated by equation PV=nRT; P-pressure of mixture in atm, V- mixture vol in liter, n- total mole number of gas mixture, T-temp of mixtue in Kelvin, R-univershal gas constant 0.082

so total pressure of gas mixture, P= nRT/V= 50X0.082X300/100= 12.3 atm [0.1 m3 = 100 liter]

c) Partial pressure of a component in a gas mixture Pi = Xi.P; Xi-mole fraction of 'i'th component, P-total pressure of the mixture.

Pnitrogen = 0.5X12.3 = 6.15 atm

PCO2 = 0.5X12.3 = 6.15 atm

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