Warming a gas at constant volume and a pressure of 1.50 atm from 27 degrees Celsius to 127 degrees Celsius gives a final pressure of

Final pressure of gas = 2 atm

Explanation:

Given data:

Initial pressure of gas = 1.50 atm

Initial temperature = 27°C

Final temperature = 127°C

Final pressure of gas = ?

Solution:

Initial temperature = 27°C (27+273 = 300 K)

Final temperature = 127°C ( 127+273 = 400 K)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1.50  atm / 300 K = P₂/400 K

P₂ = 1.50 atm × 400 K / 300 K

P₂ = 600 atm. K /300 K

P₂ = 2 atm

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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