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lyudmila [28]
2 years ago
15

If 725 ml of N² gas is at 1 atm, what volume will it have at 1.4 atm?​

Chemistry
2 answers:
nasty-shy [4]2 years ago
8 0

Answer:

<u>517.86 mL</u>

Explanation:

<u>Boyle's Law Formula</u>

  • P₁V₁ = P₂V₂

<u>Here</u> :

  • V₁ = 725 mL
  • P₁ = 1 atm
  • P₂ = 1.4 atm

<u>Solving</u>

  • V₂ = P₁V₁/P₂
  • V₂ = 1 × 725 / 1.4
  • V₂ = <u>517.86 mL</u>
alexdok [17]2 years ago
7 0

Answer:

518 mL

Explanation:

We can solve this using Boyle's Law Formula

P1V1 = P2V2

where p1 = initial pressure, p2 = final pressure, v1 = initial volume and v2 = final volume

here , the initial pressure is 1 atm and the initial volume is 725mL

we are given the final pressure 1.4 and we need to find the final volume

so we have p1v1 = p2v2

==> plug in p1 = 1 , v1 = 725 mL and p2 = 1.4

(1)(725) = (1.4)v2

==> multiply 1 and 725

725 = (1.4)(v2)

==> divide both sides by 1.4

v2 = 518

N2 would have a volume of 518mL at 1.4atm

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Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

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0.046 %

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R_{out} = \frac{A}{6000}*2000

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We can say that:

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Integration of the above linear equation =

e^{\int\limits \frac {1}{3}dt } = e^{\frac{1}{3}t

so we have:

e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A = 0.8e^{\frac{1}{3}t

\frac{d}{dt}[e^{\frac{1}{3}t}A] = 0.8e^{\frac{1}{3}t

Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C

∴ A(t) = 2.4 +Ce^{-\frac{1}{3}t

Since A(0) = 12

Then;

12 =2.4 + Ce^{-\frac{1}{3}}(0)

C= 12-2.4

C =9.6

Hence;

A(t) = 2.4 +9.6e^{-\frac{t}{3}}

A(0) = 2.4 +9.6e^{-\frac{10}{3}}

A(t) = 2.74

∴ the concentration at 10 minutes is ;

=  \frac{2.74}{6000}*100%

= 0.0456667 %

= 0.046% to three decimal places

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3 years ago
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