Question

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

Answer 1

Given:

Initial Volume $\sf (V_1)$ = 11.2 L

Initial Pressure $\sf (P_1)$ = 0.860 atm

Final Volume $\sf (V_2)$ = 15.0 L

To Find:

Final Pressure $\sf (P_2)$

Concept/Theory:

$\bf{ \underline{Boyle's \: Law} \: (Pressure - Volume \: Relationship)}$

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

$\bf{P \propto \dfrac{1}{V} \: (at \: constant \: T \: and \: n)}$

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

$\bf{PV = Constant}$

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

$\bf{P_1V_1 = P_2V_2 = Constant}$

OR

$\bf{\dfrac{P_1}{P_2} = \dfrac{V_2}{V_1}}$

Answer:

By using Boyle's Law, we get:

$\rm \longrightarrow \dfrac{0.860}{P_2} = \dfrac{15.0}{11.2} \\ \\ \rm \longrightarrow P_2 = \dfrac{11.2}{15.0} \times 0.860 \\ \\ \rm \longrightarrow P_2 = \dfrac{9.632}{15.0} \\ \\ \rm \longrightarrow P_2 = 0.642 \: atm$

$\therefore$ Final Pressure $\sf (P_2)$ = 0.642 atm

Answer 2

Answer:

0.642 atm

Explanation:

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

$P_1V_1 = P_2V_2$

Since we are finding the new pressure

$P_2 = \frac{P_1V_1}{V_2}$

From the question we have

$P_2 = \frac{11.2 \times 0.86}{15} = \frac{9.632}{15} \\ = 0.64213333...$

We have the final answer as

0.642 atm

Hope this helps you