At a temperature of 280 k, the gas in a cylinder has a volume of 20.0 liters. if the volume of the gas is decreased to 10.0 liters, what must the temperature be for the gas pressure to remain constant?

a. 140 k
b. 273 k
c. 560 k
d. 5600 k

To solve this we assume that the gas inside is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

T2 = T1 x V2 / V1

T2 = 280 x 20.0 / 10

T2 = 560 K

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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