A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes 15.0 L? (Boyle)

Answers

Answer 1

Answer:

0.642 atm

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

Since we are finding the new pressure

[tex]P_2 = \frac{P_1V_1}{V_2} \\[/tex]

From the question we have

[tex]P_2 = \frac{11.2 \times 0.86}{15} = \frac{9.632}{15} \\ = 0.64213333...[/tex]

We have the final answer as

0.642 atm

Hope this helps you

Answer 2

Given:

Initial Volume [tex] \sf (V_1) [/tex] = 11.2 L

Initial Pressure [tex] \sf (P_1) [/tex] = 0.860 atm

Final Volume [tex] \sf (V_2) [/tex] = 15.0 L

To Find:

Final Pressure [tex] \sf (P_2) [/tex]

Concept/Theory:

[tex] \bf{ \underline{Boyle's \: Law} \: (Pressure - Volume \: Relationship)}[/tex]

"At constant temperature, the pressure of a fixed amount of gas varies inversely with the volume of the gas."

[tex] \bf{P \propto \dfrac{1}{V} \: (at \: constant \: T \: and \: n)}[/tex]

It can be also stated as "At constant temperature, the product of pressure and volume of fixed amount of a gas remains constant."

[tex] \bf{PV = Constant}[/tex]

If the initial pressure and volume of a fixed amount of gas at constant temperature are [tex] \sf (P_1) [/tex] & [tex] \sf (V_1) [/tex] and final pressure of the gas is [tex] \sf (P_2) [/tex] and volume occupied is [tex] \sf (V_2) [/tex], then according to Boyle's law;

[tex] \bf{P_1V_1 = P_2V_2 = Constant}[/tex]

OR

[tex] \bf{\dfrac{P_1}{P_2} = \dfrac{V_2}{V_1}}[/tex]

Answer:

By using Boyle's Law, we get:

[tex] \rm \longrightarrow \dfrac{0.860}{P_2} = \dfrac{15.0}{11.2} \\ \\ \rm \longrightarrow P_2 = \dfrac{11.2}{15.0} \times 0.860 \\ \\ \rm \longrightarrow P_2 = \dfrac{9.632}{15.0} \\ \\ \rm \longrightarrow P_2 = 0.642 \: atm[/tex]

[tex] \therefore [/tex] Final Pressure [tex] \sf (P_2) [/tex] = 0.642 atm


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Answers

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