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expeople1 [14]
1 year ago
10

A sample of gas occupies a volume of 72.0 mL . As it expands, it does 141.2 J of work on its surroundings at a constant pressure

of 783 Torr . What is the final volume of the gas?
Chemistry
1 answer:
bagirrra123 [75]1 year ago
3 0

The final volume of the gas is  73.359 mL

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Given :

A sample gas has an initial volume of 72.0 mL

The work done = 141.2 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since, the process does 141.2 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm; we have

External Pressure Pext:

P ext = 783 torr × \frac{1 atm}{760 torr}

Pext = 1.03 atm

The work done W =  Pext V

The change in volume ΔV= \frac{W}{Pext}

ΔV = \frac{141.2 J \frac{1 L atm}{101.325 J} }{1.03 atm}

ΔV = \frac{0.0014}{1.03}

ΔV = 0.001359 L

ΔV = 1.359 mL

The initial  volume = 72.0 mL

The change in volume V is ΔV = V₂ - V₁

-  V₂ = - ΔV - V₁

multiply both sides by (-), we have:

V₂ = ΔV + V₁

     =  1.359 mL + 72.0 mL

     = 73.359 mL

Therefore, the final volume of the gas is  73.359 mL .

Learn more about volume here:

brainly.com/question/27100414

#SPJ4

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They are more stable than alkanes

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pH = 8.0

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Let's consider the balanced equation.

C₂H₄O₃ + NaOH ⇒ C₂H₃O₃Na + H₂O

The molar ratio C₂H₄O₃: NaOH: C₂H₃O₃Na is 1: 1: 1. So, when 7.2 × 10⁻⁴ moles of NaOH react completely with 7.2 × 10⁻⁴ moles of C₂H₄O₃ they form 7.2 × 10⁻⁴ moles of C₂H₃O₃Na.

The concentration of C₂H₃O₃Na is:

\frac{7.2\times 10^{-4}mol}{60.8 \times 10^{-3}L} =0.012M

C₂H₃O₃Na dissociates according to the following equation:

C₂H₃O₃Na(aq) ⇒ C₂H₃O₃⁻(aq) + Na⁺(aq)

C₂H₃O₃⁻ comes from a weak acid so it undergoes basic hydrolisis.

C₂H₃O₃⁻ + H₂O ⇄ C₂H₄O₃ + OH⁻

If we know that pKa for C₂H₄O₃ is 3.9, we can calculate pKb for C₂H₃O₃⁻ using the following expression:

pKa + pKb = 14

pKb = 14 -3.9 = 10.1

10.1 = -log Kb

Kb = 7.9 × 10⁻¹¹

We can calculate [OH⁻] using the following expression:

[OH⁻] = √(Kb.Cb)               <em>where Cb is the initial concentration of the base</em>

[OH⁻] = √(7.9 × 10⁻¹¹ × 0.012M) = 9.7 × 10⁻⁷ M

Now, we can calculate pOH and pH.

pOH = -log [OH⁻] = -log (9.7 × 10⁻⁷) = 6.0

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