Question

A certain gas is present in a 15.0 L cylinder at 2.0 atm pressure. If the pressure is increased to 4.0 atm the volume of the gas decreases to 7.5 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. Express your answers to two significant figures separated by a comma.

Answer 1

Answer:

Ki = 30atmL

Kf = 30atmL

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 15.0L

P1 (initial pressure) = 2atm

V2 (final volume) = 7.5L

P2 (final pressure) = 4atm

Ki (initial constant) =?

Kf (final constant) =?

Boyle's law states that PV = K

Ki = P1V1

Ki = 2 x 15 = 30atmL

Kf = P2V2

Kf = 4 x 7.5 = 30atmL

Since Ki and Kf are the same, the gas obey Boyle's law

Answer 2

The two constants are:

Ki = 30 atm L

Kf = 30 atm L

Given:

V₁ (initial volume) = 15.0L

P₁ (initial pressure) = 2atm

V₂ (final volume) = 7.5L

P₂ (final pressure) = 4atm

To find:

Ki (initial constant) =?

Kf (final constant) =?

According to Boyle's law:

It states that the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; i.e., in equation form,

PV = K, a constant.

Ki = P₁V₁

Ki = 2 x 15 = 30atmL

Kf = P₂V₂

Kf = 4 x 7.5 = 30atmL

Since, Ki and Kf are the same, the gas obey Boyle's law.

Find more information about Boyle's law here:

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