Polonium-210, a naturally occurring radioisotope, is an alpha emitter, with t1/2=138 d. Assume that a sample fo 210Po with a mass of 0.700 mg was placed ina 250.0-mL flask, which was evacuated, sealed, and allowed to sit undisturbed. What would the pressure be inside the flask (in mmHg) at 20 degrees Celsius after 365 days if all the alpha particles emitted has become helium atoms?

Video transcript

Welcome back everyone. Radon is a radioactive noble gas. It is formed when uranium 2 38 with a half life of 4.4, 68 years undergoes a series of nuclear decays. If a 680.4 50 mg sample of uranium 2 38 is enclosed in an empty 100 mL sealed flask. Calculate the pressure inside the flask at STP After 50,000 years. Assuming that all the pressure inside the flask is due to the radon gas. So we're going to recall that whenever we have radioactive decay we'll be following first order kinetics always. So we're going to recall our rate law where and what will write rate law out first. So the following rate law where we take the natural log of our number of radioactive isotopes present after the amount of time for decay passes. So in this case, after 50,000 years Divided by an zero, which represents the amount of radioisotope we have before 50,000 years passes. So before decay begins, this is all set equal to negative one times are decay constant K, which K is then multiplied by the amount of time that passes T and according to the prompt T is equal to 50,000 years. So we don't have our decay constant, but we should recall that to calculate our decay constant, we would find this by taking the natural log Of the value two divided by our half life. And in the prompt were given the half life of our isotope as This value here, 4.468 times 10 to the ninth power years. So we'll plug that in in the denominator. We have 4.4 68 times 10 to the ninth power years as our half life of uranium 2 38. And then when we take the Ln of two, in our numerator will have a value of 20.6 43. So in our calculators, when we take, when we type in this quotient, we'll have a result equal to 1.5510295 times 10 to the negative 10th power in verse years. Now, with this value for our decay constant, we want to begin by finding our value for N and zero specifically, which is our initial amount of uranium 2 38 atoms. And so according to the prompts were actually given the sample size of this uranium 238 were given that sample being . mg of uranium 238. And so we're going to convert using dimensional analysis where milligrams will be in the denominator and grams will be in the numerator and recall that our prefix milli tells us we have 10 to the negative third of our base unit millie until this leads us to grams of uranium 2 38. After we cancel out milligrams, now going from grams, we're going to go into moles. So we'll have moles in the numerator. And in this case, we're going to recognize that because we have an isotope recall that for any isotope, the number given in the left hand exponents is the mass number. And so we can understand this as 238 g of uranium 2 38 equivalent to one mole. This allows us to cancel out grams. Now we can go from moles of uranium 2 two atoms of uranium 2 38 as our final unit where we would recall avocados number which tells us that for one mole of uranium 2 we have an equivalent of six point oh 22 times 10 to the 23rd power atoms. And so not canceling out moles were left with atoms where when we take the entire product of all of our quotients, we should come up with a result equal to a value of 1.1386134 times 10 to the 18th power atoms of uranium 38 present Before 50,000 years. So we'll just say initially. So this is all equal to N0. Now, with this value, we're going to find our value in the numerator of our rate law being N. So the number of so N is the amount of 2 38 uranium atoms after the decay. So after 50,000 years, so solving for this, we're going to utilize the rate law. So we'll have the natural log of our numerator end which we're trying to find divided by the end zero value, which we just solved for as 1.1386134 times 10 to the 18th power atoms Of Uranium 2 38. This is set equal to negative one times our decay constant, which we found above as 1.5510, 295 times 10 to the negative 10th power inverse years. And this is then multiplied by the amount of time that passes given in the prompt as 50, years. So we're going to begin by taking the product of our right hand side. So we'll still have the same left hand side. So I'm just going to copy and paste that instead of taking the time to write everything. So again, our left hand side is the same and the product of our right hand side is going to equal a value of negative 7.75514 times 10 to the negative six power. And as far as the units notice that will be able to cancel out inverse years with years when we multiply them So now we have no units on the right hand side. And now we're just going to get rid of the log term next by recalling that we can cancel out the natural log using Mueller's number. So making both sides and exponents Mueller's number. And so this will cancel L N on the left hand side leaving us with just our fraction. So we'll just copy and paste that where and sorry about that. So pasting this again, we won't need our parentheses anymore. But this will be set equal to the result of the right hand side where we should come up with a value of zero point 99999224489. And now with this, we want to get rid of our denominator on the left hand side. So we're gonna multiply both sides by 1. times 10 to the 18th power atoms of uranium 2 on both sides, it'll cancel out on the left hand side and on the right hand side, we multiply again by the same thing. So let me just make sure I'm writing this properly 1.1386134 times 10 to the 18th power atoms of uranium 2 38. So taking the product on the right hand side will find that our number of radioisotope we have after 50,000 years of decay occurs is equal to a value of 1.1386046 times 10 to the 18th power atoms of uranium 38 that remain Again after 50,000 years. And so now that we have what we have of our radio isotope before the decay. And after the decay, we want to figure out how many atoms actually are decayed total. So to find our atoms of uranium 238 that are decayed in total, We would just take the difference between N0 and N. And so plugging in our value for N0, which we found above as 1. 6, 4 times 10 to the 18th power Atoms of uranium 2 38 and subtracting this from what we just solved for as what we have left remaining after 50, years, 1.1386046 times 10 to the 18th power Atoms of Uranium 38. This difference will result in a value of our total atoms decayed equal to 8.8 times 10 to the 12th power uranium to 38 atoms decayed. And now with this amount of Adam's decade, we're going to recall that our flask containing the decayed uranium 238 forms rate on gas. So radon gas is formed from our isotope. And so we can understand that Our uranium 2 38 that is decayed should be equal to R rated on atoms and we'll just say RN our raid on Adams produced. And so we can say, therefore, we have 8.8 times 10 to the 12 power Reagan atoms. But because we need to figure out the pressure of the flash we want to get from atoms to mold. So we're going to utilize dimensional analysis again. So scrolling down for more room, we're going to go from Right on Adams. So again, we have 8.8 times 10 to the 12 power rate on Adams, we want to cancel that out and get some old. So we'll have atoms of radon in the denominator and moles of radon in the numerator where we would utilize again, avocados number which tells us we have an equivalent of six point oh 22 times 10 to the 23rd power of atoms of radon equivalent to one mole of radon. This allows us to get rid of atoms were left with moles and we should come up with our moles of radon equal to 1.4613, 085 times 10 to the negative 11th power moles of Reagan. Now, with this value of moles of our radon gas, we're going to recall the formula PV equals and R T to sulfur pressure. Now, according to the prompt, we have a volume of the flask equal to mL and recall that R gas constant R Utilizes the value 0.08, and we have units of leaders times A T M divided by moles times Kelvin. And because we want everything to cancel out easily. When we use our formula here, we're going to want to convert this volume from middle leaders to leaders. So we're called that again are prefix Milli tells us we have 10 to the negative third power of our base unit leaders. So canceling out middle leaders were left with leaders and we'll have a volume and leaders equal to 0.100 leaders. Now with our units mapped out, let's plug in what we know, but we need to isolate for pressure. So we're going to divide both sides by volume. And what we'll have is that our pressure is equal to our most of our gas of Reagan times the gas constant, R times the temperature in Kelvin divided by V volume. Recalled that the prompt mentions that we're at S T P. And so therefore, our temperature is equal to 2 73.15 Kelvin because we want our units of Kelvin to cancel out from our gas constant. So let's plug in everything we know into our formula. Pressure is equal to our moles of radon gas which we just found above as one point sorry 1. times 10 to the negative 11th power moles of Reagan. This is multiplied by R gas constant R which we just recalled as 0.8206 leaders times A T M's divided by moles times Kelvin. And then this is multiplied by our temperature standard temperature. In Kelvin being to 73.15. Kelvin then dividing by the volume we just found as 0.100 liters. We can cancel out leaders with leaders in the numerator, Kelvin with Kelvin here and then we can get rid of moles with moles here. Leaving us with A T M as our final unit of pressure, which is what we want. So when we take the result of the quotient in our calculators, we should come up with a value equal to 3. or sorry. It's just 75477556 times 10 to the negative ninth power A T M of radon gas produced inside the flask after 50,000 years of decay of uranium 2 38 occurs. Recall that our prompt utilizes three sig figs as our minimum amount. So our final answer is going to be just 3.27 times 10 to the negative ninth power A T M's of Reagan. So this final answer highlighted in yellow is going to complete this example as our radon gas produced and corresponds to choice C in the multiple choice. I hope that everything I went through was clear. If you have any questions, please leave them below and I'll see everyone in the next practice video.