Aballoon filled with gas expands in volume by 2.0 l against an external pressure of 0.93 atm. if the balloon has absorbed 636 j of heat from the surroundings, what is the change in the balloon’s internal energy? (note: 1 l-atm = 101.3 j)

447.58 J

Explanation:

The change in internal energy is the difference between the heat absorbed and the work done due to gas expansion. The following variables are given in the problem:

Change in volume = ΔV = 2.0 L

External pressure = P = 0.93 atm

Work done due to gas expansion = W = P*ΔV  = 2*0.93 = 1.86 L-atm

We can convert the work done from L-atm to J by using the conversion unit given in the question: 1 L-atm = 101.3 J. Thus:

Work done = 1.86*101.3 = 188.418 J

Change in Internal energy (ΔU) = heat absorbed (Q) - work done (W)

The heat absorbed is equivalent to 636 J. thus, the internal energy is:

ΔU = 636 - 188.418 = 447.58 J

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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