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

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

it means there are 3 atoms of oxygen

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.