Acertain gas is present in a 13.0 l cylinder at 1.0 atm pressure. if the pressure is increased to 2.0 atm , the volume of the gas decreases to 6.5 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.

The gas obeys the Boyle's law

Explanation:According to Boyle's law, the volume of a fixed mass of a gas and the pressure are inversely proportional at constant absolute temperature.That is; Therefore, , where k is a constant At varying volume and pressures while keeping absolute temperature constant; k = P1V1 =P2V2

In this case, we are given;

Initial Volume of 13.0 L at initial pressure of 1.0 atm

New volume of 6.5 L at new pressure of 2.0 atm

But, K = PV

Therefore,

k1 = P1V1

   = 1.0 atm × 13.0 L

  = 13 atm.L

k2 = P2V2

    = 2.0 atm × 6.5 L

    = 13 atm.L

Thus, k1=k2

Thus, the gas obeys the Boyle's law

D

Explanation:

We can use the mole ratio to calculate the partial pressure. The total number of moles is 0.2 + 0.2 + 0.1 = 0.5 moles

Now, we know that the mole fraction of the argon gas would be 0.2/0.5

The partial pressure is as follows. To calculate this, we simple multiply the number of moles by the total pressure.

0.2/0.5 * 5 = 1.0/0.5 = 2.00atm

D