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Sindrei [870]
3 years ago
13

a balloon has a pressure of 104.5 kpa and a volume of 3.4L. If the pressure is reduced to 42.7 kpa, what will the volume change

to?​
Chemistry
2 answers:
Nuetrik [128]3 years ago
7 0

Answer:

The answer to your question is 8,32 L

Explanation:

Data

Pressure 1 = 104.5 kPa

Volume 1 = 3.4 L

Pressure 2 = 42.7 kPa

Volume 2 = ?

Formula

- To solve this problem use the Boyle's law.

       P1V1 = P2V2

- Solve for V2

       V2 = P1V1 / P2

- Substitution

       V2 = (104.5)(3.4) / 42.7

-Simplification

       V2 = 355.3 / 42.7

- Result

        V2 = 8.32 L            

Nookie1986 [14]3 years ago
3 0

Answer:

The new volume is 8.3 L

Explanation:

Step 1: Data given

Initial pressure of balloon = 104.5 kPa = 1.0313348 atm

Volume of the balloon is 3.4 L

The pressure is reduced to 42.7 kPa = 0.4214162 atm

Step 2: Calculate the new volume

p1*V1 = p2 *V2

⇒p1 = the initial pressure in the balloon = 1.0313348 atm

⇒V1 = the initial volume = 3.4 L

⇒p2 = the reduced pressure = 0.4214162 atm

⇒V2 = the new volume

1.0313348 atm * 3.4 L = 0.4214162 * V2

V2 = 8.3 L

The new volume is 8.3 L

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5 0
2 years ago
41 to
tia_tia [17]

Answer:

d = 0.9 g/L

Explanation:

Given data:

Number of moles = 1 mol

Volume = 24.2 L

Temperature = 298 K

Pressure = 101.3 Kpa (101.3/101 = 1 atm)

Density of sample = ?

Solution:

PV = nRT     (1)

n = number of moles

number of moles = mass/molar mass

n = m/M

Now we will put the n= m/M in equation 1.

PV = m/M RT   (2)

d = m/v

PM = m/v RT ( by rearranging the equation 2)

PM = dRT

d = PM/RT

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d = 1 atm × 20.1798 g/mol / 0.0821 atm. L/mol.K × 273K

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d = 0.9 g/L

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