Question

A certain gas is present in a 10.0 L cylinder at 4.0 atm pressure. If the pressure is increased to 8.0 atm the volume of the gas decreases to 5.0 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

Hence , Ki = Kf

The gas is obeying the Boyle's law.

Step-by-step explanation:

Given data:

The initial volume of the cylinder(in litres) =
`V_1` = 12.0 L

The initial pressure(in atmospheric pressure) =
`P_1`= 4.0 atm

The final pressure(in atmospheric pressure) =
`P_2` = 8.0 atm

The final volume of the cylinder(in litres) =
`V_2` = 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 = (k)/(V)`

Where,

P = Pressure

V = Volume of the gas

k = Boyle's constant

According to the Boyle's law,

`P_(1) = (k_(i))/(V_(1))`

`=> k_(i) = P_(1)V_(1)`

Plug-in the values in the above equation, you would get:

`k_(i) = 4.0 * 10.0 = 40`

The final pressure(in atmospheric pressure) = = 8.0 atm

The final volume of the cylinder(in litres) == 5.0 L

The Boyle's constant =
`k_(f)` = ?

According to the Boyle's law,

`P_(2) = (k_(f))/(V_(2))`

`=> k_(f) = P_(2)V_(2)`

Plug-in the values in the above equation, you would get:

`k_(f) = 8.0 * 5.0 = 40`

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) = 40\\k_(f) = 40`

Therefore,

40 = 40

Hence , Ki = Kf

The gas is obeying the Boyle's law.