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Art [367]
2 years ago
8

A gas sample has a volume of 30.0 mL at a pressure of 1.53 atm. If the volume increases to 50.1 mL and the temperature remains c

onstant, the new pressure in atms will be
Chemistry
1 answer:
Tems11 [23]2 years ago
6 0

At constant temperature, if the volume of the sample of gas increases to the given value, the pressure decreases to 0.92atm.

<h3>Boyle's law</h3>

Boyle's law simply states that "the volume of any given quantity of gas is inversely proportional to its pressure as long as temperature remains constant.

Boyle's law is expressed as;

P₁V₁ = P₂V₂

Where P₁ is Initial Pressure, V₁ is Initial volume, P₂ is Final Pressure and V₂ is Final volume.

Given the data in the question question;

  • Initial volume of the gas V₁ = 30.0mL = 0.03L
  • Initial pressure of the gas P₁ = 1.53atm
  • Final volume of the gas V₂ = 50.1mL = 0.0501L
  • Final pressure of the gas P₂ = ?

We substitute our given values into the expression above to determine the new pressure.

P₁V₁ = P₂V₂

P₂ = P₁V₁ / V₂

P₂ = ( 1.53atm × 0.03L ) / 0.0501L

P₂ = 0.0459Latm / 0.0501L

P₂ = 0.92atm

Therefore, at constant temperature, if the volume of the sample of gas increases to the given value, the pressure decreases to 0.92atm.

Learn more about Boyle's law here: brainly.com/question/1437490

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2 years ago
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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch
Salsk061 [2.6K]

Answer:

kp= 3.1 x 10^(-2)

Explanation:

To solve this problem we have to write down the reaction and use the ICE table for pressures:

                                2SO2      +        O2         ⇄              2SO3

Initial                      3.4 atm           1.3 atm                         0 atm

Change                    -2x                    - x                                + 2x

Equilibrium            3.4 - 2x            1.3 -x                          0.52 atm

In order to know the x value:

2x = 0.52

x=(0.52)/2= 0.26

                               2SO2             +          O2              ⇄              2SO3

Equilibrium        3.4 - 0.52                1.3 - 0.26                     0.52 atm

Equilibrium        2.88 atm                 1.04 atm                      0.52 atm

with the partial pressure in the equilibrium, we can obtain Kp.

Kp=\frac{PSO3^2}{PSO2^2 PO2}=\frac{(0.52)^2}{(2.88)^2(1.04)}=0.03135

8 0
3 years ago
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