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9966 [12]
3 years ago
8

You just purchased a 400-L rigid tank for a client who works in the gas industry. The tank is delivered pre-filled with 3 kg of

air at a temperature of 25C. Assuming an atmospheric pressure of 98 kPa, determine the pressure reading if you connected a pressure gage to the tank.
Engineering
1 answer:
solniwko [45]3 years ago
6 0

Answer:

the pressure reading when connected a pressure gauge is 543.44 kPa

Explanation:

Given data

tank volume (V) = 400 L i.e 0.4 m³

temperature (T) =  25°C  i.e. 25°C + 273 = 298 K

air mass (m)  = 3 kg

atmospheric pressure  = 98 kPa

To find out

pressure reading

Solution

we have find out pressure reading by gauge pressure

i.e. gauge pressure = absolute pressure - atmospheric pressure

first we find absolute pressure (p) by the ideal gas condition

i.e pV = mRT

p = mRT / V

p = ( 3 × 0.287 × 298 ) / 0.4

p = 641.44 kPa

so

gauge pressure = absolute pressure - atmospheric pressure

gauge pressure = 641.44 - 98

gauge pressure = 543.44 kPa

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) An ammonia nitrogen analysis performed on a wastewater sample yielded 30 mg/L as nitrogen. If the pH of the sample was 8.5, de
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Answer:

NH_4^+ = 2.5 mg/lt

Explanation:

Given data:

Ammonia Nitrogen 30 mg/L

pH = 8.5

-log[H +] = 8.5

[H +] = 10^{-8.5}

NH_4 ^{+} ⇄ H^{+} + NH_3

Rate constant is given as

K_a = \frac{[H^{+}] [NH_3]}{NH_4^{+}} ...........1

K_a = 5.6 \times 10^{-10}

Total ammonia as NItrogen is given as 30 mg/l

\%NH_4^{+} = \frac{ [NH_4] \times 100}{[NH_4^{+}] + [NH_3]}

                    = \frac{100}{\frac{NH_4^+}{NH_4^+} +\frac{NH_3^+}{NH_4^+}}

                    = \frac{100}{1+ \frac{NH_3^+}{NH_4^+}} .....2

from equation 1 we have

\frac{NH_3^+}{NH_4^+} =\frac{K_a}{[H^+]} = \frac{5.6\times 10^{-10}}{10^{8.5}}

plug this value in equation 2 we get

\%NH_4^{+} = 84.96 \%

Total ammonia as N = 30 mg/lt

NH_4^+ = \frac{84.96}{100} \times 30 = 25.5 mg/lt

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