A weather balloon is partially filled with helium to allow for...

Question

A weather balloon is partially filled with helium to allow for expansion at high altitudes. At STP, a weather balloon is filled with enough helium to give a volume of 25.0 L. At an altitude of 30.0 km and –35 ⁰C, it has expanded to 2460 L. The increase in volume causes it to burst and a small parachute returns the instruments to Earth.
b. What is the final pressure, in millimeters of mercury, of the helium inside the balloon when it bursts?

b. What is the final pressure, in millimeters of mercury, of the helium inside the balloon when it bursts?

Accepted Answer

Welcome back, everyone at S T P A container is filled with gas to an initial volume of 3.50 liters. The temperature was increased to 115 Celsius and the container was allowed to expand to a volume of 50 liters. Calculate the final pressure in milliliters of mercury of the gas in the container. So let's begin by recalling our combined gas law in which we can use to calculate our final pressure. Recall that our combined gas law outlines our initial pressure times our initial volume divided by initial temperature conditions, which is set equal to our final pressure, times final volume divided by our final temperature conditions. So in short, that is P one times V one over T one, equal to P two times V two over T two. We need to isolate for P two, which is our final pressure. So we're going to multiply both sides by our denominator T two and divide both sides by our numerator V two. And in doing so, we would cancel out those terms on the right, which would now leave us with P one times V, one times T two, divided by T one times V two and this is all equal to our final pressure P two. So now we're going to outline our given conditions beginning with our initial conditions. So our initial pressure that is given to us is S T P. So for at S T P then recall that standard pressure is going to be one point oh ATM or just one ATM. Next, we're given our initial volume. So for our initial volume, we have a value of 3.50 liters. And then lastly, we're given our initial temperature which at S T P. Therefore, it is going to be 273.15. Kelvin. We're also given a bit of our final conditions. So noting those down, now we need to find P two, we do not know that. However, we're given our final volume equal to 50.0 liters and we're also given our final temperature T two, which we're told increases to 115 Celsius. So we would take our initial temperature to 73.15 Kelvin plus the final temperature of 100 or the increase in temperature sorry of 115 point we actually just 115 degrees Celsius. So taking the sum between these two, we would have a total of 388. Kelvin. So now with most of our variables outlined, let's plug in what we know to solve for our final pressure P two. So plugging in our values for initial pressure, we said we have 1. ATM which is multiplied by our initial volume of 3. liters and then multiplied by our final temperature which we just determined to be 388.15. Kelvin. In our denominator, we have our initial temperature, which we determined above is for standard temperature 73.15, Kelvin, which is multiplied by our final volume, which we noted down as 50.0 liters. So noticing that we can cancel our units of leaders with leaders in the denominator, Kelvin, with Kelvin in the denominator, we're left with ATM as our final units which will need to convert at our final step. However, in simplifying this, this is equal to P two. And in our next line, the result we'll get from our quotient is going to be a value of 1358.53 ATM in our numerator which is divided by 13, 0.5 set equal to our final pressure and then dividing our quotient, we would get a value of 0.99471 ATM equal to our final pressure. So now with this final pressure outlined, we're going to need to convert this to millimeters of mercury. So in doing so, we would have our final pressure P two equal to 0.99471 ATM multiplied by our conversion factor to go from ATM in the denominator to millimeters of mercury in the numerator where we should recall from our textbooks that one ATM has an equivalence of 760 millimeters of mercury in the numerator canceling out our units of ATM. Now we're left with millimeters of mercury for units of our final pressure P two. And as a result, we will simplify so that we get our answer as 75.5, 98 millimeters of mercury as the final pressure in our container of gas. Now, based on our operation of multiplying and dividing here, our lowest number of sig figs and actually just to make a correction going to our initial pressure were given or we were determined that at S T P standard temperature and pressure, we have a pressure of 1.0 ATM. This is a measured and constant value with three significant figures. So if our initial pressure has three significant figures, then in our final calculation, we should have three significant figures for our pressure surrounding this to about three sigfig. Our final answer is going to be 75.6 millimeters of mercury. And this would be our final answer as our value for P two. Our final pressure inside of our container filled with gas. So I hope that this made sense and let us know if you have any questions.

Key Concepts

Ideal Gas Law

Standard Temperature and Pressure (STP)

Gas Expansion and Pressure Changes

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