A sample of xenon gas at a pressure of 817 mm Hg and a temperature of 66 °C, occupies a volume of 13.9 liters. If the gas is heated at constant pressure to a temperature of 89 °C, the volume of the gas sample will be 14.84 L.

The value of heat transfer in xenon gas during heating at constant pressure  Q = 0.254  KJ

Explanation:

Initial pressure = 109 k pa = Final pressure

Initial temperature = 66°c = 339 K

Initial volume = 13.9 liter = 0.0139

Final temperature = 362 K

Final volume = 14.84 liter

Specific heat for xenon at constant pressure = 0.158

Gas constant  for xenon (R)= 0.063

From ideal gas equation

= m R ------ (1)

⇒ 109 × 0.0139 = m × 0.06333 × 339

⇒ m = 0.07 kg

This is the mass of xenon gas.

heat transfer is given by

Put all the value sin above equation we get

Q = 0.07 × 0.158 × (362 - 339)

Q = 0.254  KJ

This is the value of heat transfer in xenon gas during heating at constant pressure.

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.

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.