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TEA [102]
4 years ago
5

Define a function ComputeGasVolume that returns the volume of a gas given parameters pressure, temperature, and moles. Use the g

as equation PV = nRT, where P is pressure in Pascals, V is volume in cubic meters, n is number of moles, R is the gas constant 8.3144621 ( J / (mol*K)), and T is temperature in Kelvin.
Chemistry
1 answer:
andre [41]4 years ago
7 0

Answer: A function can be defined as a relation in which one thing is dependent on another for its value.

Explanation: Given R = 8.314J/mol*k

PV = nRT

V = nRT/ P

V = 8.314RT / P (cm^3)

Volume of gas =(( 8.314 * R* T) / P ) cm3

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A power plant is driven by the combustion of a complex fossil fuel having the formula C11H7S. Assume the air supply is composed
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(a) 4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 20.68N_2;

(b) 4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2;

(c) 23 900 kg air; (d) air:fuel = 10.2; (e) air:fuel = 12.2:1

(a) <em>Balanced equation including N_2 from air</em>  

The balanced equation <em>ignoring</em> N_2 from air is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2  

Moles of N_2 =55 mol O_2 × (3.76 mol N_2/1 mol O_2) = 206.8 mol N_2  

<em>Including</em> N_2 from air, the balanced equation is  

4C_11 H_7S + 55O_2 → 44CO_2 + 14H_2O + 4SO_2 + 206.8N_2  

(b) <em>Balanced equation for 120 % stoichiometric combustion</em>  

Moles of O_2 = 55 mol O_2 × 1.20 = 66.00 mol O_2  

Excess moles O_2 = (66.00 – 55) mol O_2 = 11.00 mol O_2  

Moles of N_2 = 66.00 mol O_2 × (3.76 mol N_2/1 mol O_2) = 248.2 mol N_2  

The balanced equation is

4C_11 H_7S + 66O_2 → 44CO_2 + 14H_2O + 4SO_2 + 248.2N_2 + 11O_2

(c) <em>Minimum mass of air</em>  

Moles of O_2 required = 1700 kg C_11 H_7S

× (1 kmol C_11 H_7S/185.24 kg C_11 H_7S) × (55 kmol O_2/4 kmol C_11 H_7S)

= 126.2 kmol O_2  

Mass of O_2 = 126.2 kmol O_2 × (32.00 kg O_2/1 kmol O_2) = 4038 kg O_2  

Mass of N_2 required = 126.2 kmol O_2 × (3.76 kmol N_2/1 kmol O_2)

× (28.01 kg N_2/1 kmol N_2) = 13 285 kg N_2  

Mass of air = Mass of N_2 + mass of O_2 = (4038 + 13 285) kg = 17 300 kg air  

(d) <em>Air:fuel mass ratio for 100 % combustion</em>  

Air:fuel = 17 300 kg/1700 kg = <em>10.2 :1 </em>

(e) <em>Air:fuel mass ratio for 120 % combustion </em>

Mass of air = 17 300 kg × 1.20 = 20 760 kg air  

Air:fuel = 20 760 kg/1700 kg = 12.2 :1  

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