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Stolb23 [73]
3 years ago
12

The exhaust gas from an automobile contains 1.5 percent by volume of carbon monoxide. What is the concentration of CO in mg/m' a

t 25°C and 1 atm pressure? What is the concentration in mg/m' in the exhaust pipe if it is at 200°C and 1.1 atm of pressure?
Chemistry
1 answer:
FrozenT [24]3 years ago
5 0

Answer:

(a) 17,178 mg/m3

(b) 11,625 mg/m3

Explanation:

The concentration of CO in mg/m3 can be calculated as

C (mg/m3) =(P/RT)*MW*C_{ppm}

For standard conditions (1 atm and 25°C), P/RT is 0.0409.

Concentration of 1.5% percent by volume of CO is equivalent to 1.5*10,000 ppm= 15,000 ppm CO.

The molecular weigth of CO is 28 g/mol.

(1) For 25°C and 1 atm conditions

C=(P/RT)*MW*C_{ppm}\\\\C=0.0409*28*15,000=17,178

(b) For 200°C and 1.1 atm,

P/RT=0.0409*(P/P_{std})*(T_{std}/T)\\P/RT=0.0409*(1.1atm/1atm)*(273+15K/273+200K)=0.0277

Then the concentration in mg/m3 is

C=(P/RT)*MW*C_{ppm}\\\\C=0.0277*28*15,000=11,625

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The given function is:

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or

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<span>The maxima point is obtained by taking the 1st derivative of the function then equating dP / di = 0:</span>

dP / di = 120 (i^2 + i + 9)^-1 + (-1) 120 i (i^2 + i + 9)^-2 (2i + 1)

setting dP / di =0 and multiplying whole equation by (i^2 + i + 9)^2:

0 = 120 (i^2 + i + 9) – 120i (2i + 1)

Dividing further by 120 will yield:

i^2 + i + 9 – 2i^2 – i = 0

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<span>i = 3      (ANSWER)</span>

Therefore P is a maximum when i = 3

Checking:

P = 120 * 3 / (3^2 + 3 + 9)

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Answer:

This question is incomplete; the complete part/options is:

A) They are precise, but not accurate.

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