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dlinn [17]
3 years ago
10

A cylinder containing 20.0 L of compressed nitrogen is connected to an empty (evacuated) vessel with an unknown volume. The gas

pressure in the cylinder starts at 25 atm and drops to 2 atm without a change in temperature. Determine the volume of the vessel.
Chemistry
1 answer:
Nonamiya [84]3 years ago
3 0

Answer:

The new volume of the vessel is 250 L

Explanation:

Step 1: Data given

The volume of the cylinder = 20.0 L

The initial pressure of the gas = 25 atm

The pressure drops to 2 atm

Step 2: Calculate the volume of the vessel

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 25 atm

⇒with V1 = the initial volume = 20.0 L

⇒with P2 = the final pressure = 2 atm

⇒with V2 = the final volume = TO BE DETERMINED

25 atm* 20.0 L = 2.0 atm * V2

V2 = 250 L

The new volume of the vessel is 250 L

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3 years ago
The structural formula of 2-methyl-propane<br>-2-ol​
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8 0
3 years ago
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/m
koban [17]

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.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

R_{in} = \frac{0.04}{100}*2000

R_{in} = 0.8

The rate-out

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

R_{out} = \frac{A}{3}

We can say that:

\frac{dA}{dt}= 0.8-\frac{A}{3}

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\frac{dA}{dt} +\frac{A}{3} =0.8

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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