Question

A certain gas is present in a 12.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas decreases to 6.0 L . Find the two constants ki, the initial value of k, and kf, the final value of k, to verify whether the gas obeys Boyle’s law by entering the numerical values for ki and kf in the space provided.

Answer 1

There are 3 parts in this question:
1) To find the initial Boyle's constant $k_{i}$
2) To find the final Boyle's constant $k_{f}$
3) To verify whether gas is obeying Boyle's law or not

Given data:
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm

The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L

First you need to know what Boyle's law is:
Boyle's law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature.

The Mathematical form of Boyle's law is:
$P = \frac{k}{V}$

Where,
P = Pressure
V = Volume of the gas
k = Boyle's constant

Now let's solve aforementioned parts one by one:
1. 
The initial volume of the cylinder(in litres) = $V_{i}$ = 12.0 L
The initial pressure(in atmospheric pressure) = $P_{i}$ = 4.0 atm
The Boyle's constant = $k_{i}$ = ?

According to the Boyle's law,

$P_{i} = \frac{k_{i}}{V_{i}}$

=> $k_{i}$ =  $P_{i}V_{i}$
Plug-in the values in the above equation, you would get:
$k_{i}$ = 4.0 * 12.0 = 48

Ans-1) $k_{i}$ = 48

2.
The final pressure(in atmospheric pressure) = $P_{f}$ = 8.0 atm
The final volume of the cylinder(in litres) = $V_{f}$ = 6.0 L
The Boyle's constant = $k_{f}$ = ?

According to the Boyle's law,

$P_{f} = \frac{k_{f}}{V_{f}}$

=> $k_{f}$ =  $P_{f}V_{f}$
Plug-in the values in the above equation, you would get:
$k_{f}$ = 8.0 * 6.0 = 48

Ans-2) $k_{f}$ = 48

3.
In order to verify Boyle's law, the initial Boyle's constant should be EQUAL to the final Boyle's constant, meaning:

$k_{i}$ = $k_{f}$

Since,
$k_{i}$ = 48
$k_{f}$ = 48

Therefore,
48=48.

Ans-3) Hence proved: The gas IS obeying the Boyle's law.

-i

Answer 2

The initial value of constant ${{\text{k}}_1}$ is 48.0 atmL and the final value of constant ${{\text{k}}_2}$  is 48.0 atmL. This proves that Boyle's lawis obeyed by gas.

Further explanation:

Boyle’s law:

It is an experimental gas law that describes the relationship between pressure and volume of the gas. According to Boyle's law, the volume of the gas is inversely proportional to the pressure of the system, provided that the temperature and the number of moles of gas remain constant.

If the temperature and number of moles of gas are constant then the equation (1) will become as follows:

${\text{PV}} = {\text{k}}$                 ……(2)

Here, k is a constant.

Or it can also be expressed as follows:

${{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}$    ……(3)

Here,

${{\text{P}}_1}$ is the initial pressure.

${{\text{P}}_2}$ is the final pressure.

${{\text{V}}_1}$ is the initial volume.

${{\text{V}}_2}$ is the final volume.

Boyle'slaw for the initial condition of gas can be written as,

${{\text{P}}_1}{{\text{V}}_1}={{\text{k}}_1}$                                   …… (4)

Substitute 4.0 atm for ${{\text{P}}_1}$  and 12.0 L for ${{\text{V}}_1}$  in equation (4).

$\begin{aligned}\left( {4.0{\text{ atm}}}\right)\left({12.0{\text{ L}}}\right)&= {{\text{k}}_1}\hfill\\48.0{\text{ atm}}\cdot{\text{L}}&= {{\text{k}}_1}\hfill\\\end{aligned}$

Boyle's law for the final condition of gas can be written as,

${{\text{P}}_2}{{\text{V}}_2} = {{\text{k}}_2}$                                   …… (5)

Substitute 8.0 atm for ${{\text{P}}_2}$  and 6.0 L for ${{\text{V}}_2}$  in equation (5).

$\begin{aligned}\left( {8.0{\text{ atm}}}\right)\left({6.0{\text{ L}}}\right)&={{\text{k}}_2}\hfill\\48.0{\text{ atm}}\cdot{\text{L}}&={{\text{k}}_2}\hfill\\ \end{aligned}$

Since the value of ${{\text{k}}_1}$ and ${{\text{k}}_2}$  is equal in both cases thus this gives,

 ${{\text{P}}_1}{{\text{V}}_1} = {{\text{P}}_2}{{\text{V}}_2}$

Hence, it is proved that Boyle's law is obeyed by the given gas.

Learn more:

1. Law of conservation of matter states: brainly.com/question/2190120

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas of equation

Keywords: Boyle's law, volume, temperature, pressure, volume pressure relationship, constant temperature, relationship, V inversely proportional to P, ideal gas, ideal gas equation number of mole and moles.