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Nina [5.8K]
3 years ago
8

An air compressor draws in 1,200 cf of free air at a gauge pressure of 0 psi and a temperature of 70°F. The air is compressed to

a gauge pressure of 90 psi at a temperature of 125°F. The atmospheric pressure is 12.20 psi. Determine the volume of air after it is compressed.
Chemistry
1 answer:
lubasha [3.4K]3 years ago
7 0

Answer:

The final volume of air after compression: V₂ = 4477.63 L = 158.12 cf

Explanation:

Given: Initial gauge pressure of the gas = 0 psi

Initial absolute pressure of the gas: P₁ = gauge pressure + atmospheric pressure = 0 psi + 12.20 psi = 12.20 psi

Initial Temperature = 70°F

⇒ T₁ = (70°F − 32) × 5/9 + 273.15 = 294.26 K

Initial volume of the gas: V₁ = 1200 cf = 1200 × 28.317  = 33980.4 L    (∵ 1 cf ≈ 28.317  L)

Final gauge pressure of the gas = 90 psi

Final absolute pressure of the gas: P₂ = gauge pressure + atmospheric pressure = 90 psi + 12.20 psi = 102.2 psi

Final Temperature: T₂ = 125°F

⇒ T₂ = (125°F − 32) × 5/9 + 273.15 = 324.82 K

Final volume of the gas: V₂ = ? L      

<u>According to the </u><u>Combined gas law</u><u>:</u>

\frac{P_{1}\times V_{1}}{T_{1}} = \frac{P_{2}\times V_{2}}{T_{2}}

→ V_{2} = \frac{P_{1}\times V_{1}\times T_{2}}{T_{1}\times P_{2}}

→ V_{2} = \frac{12.20 psi\times 33980.4 L\times 324.82 K}{294.26 K\times 102.2 psi}

→ V_{2} = 4477.63 L = 158.12 cf

<u>Therefore, the final volume of air after compression: </u><u>V₂ = 4477.63 L = 158.12 cf</u>

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Yes, Mass is conserved.

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Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be
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The equation of the titration between the benzoic acid and NaOH is:

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a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

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\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles  

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles

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C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺  

0.14 - x                            x                x

Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}

Ka = \frac{x*x}{0.14 - x}

6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0   (2)  

By solving equation (2) for x we have:          

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52

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C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻     (3)

The number of moles of C₆H₅CO₂⁻ is:

\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles

The volume of NaOH added is:

V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L

The concentration of C₆H₅CO₂⁻ is:

C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻  

0.14 - x                              x               x

Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}

(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0

(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0

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x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

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From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:                              

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pOH = -log([OH^{-}]) = -log(0.067) = 1.17

pH = 14 - pOH = 12.83

I hope it helps you!    

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