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Ch.10 - Gases: Their Properties & Behavior

Chapter 10, Problem 97

At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 580 mph?

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Welcome back everyone. We need to calculate the temperature in degrees Celsius, wherein the main speed of flooring gas molecules is the same as a private jet traveling at 650 MPH. So we're going to first recall our root mean square velocity or speed formula where we would take the square root of three, multiplied by r. Gas constant R multiplied by r temperature in kelvin divided by the mass of our gas by its molar mass specifically, which is why we have a capital M here. And we're going to recognize that if we use this formula, we would get temperature in kelvin and we would recall that our kelvin temperature is really equal to our degrees Celsius temperature plus 273. So we would just take the difference to end up with our final answer in degrees Celsius. Now we're going to begin by recognizing the units of our root, mean squared agreed formula where our root mean square speed is in units of meters per second. We have our gas constant R which involves units of Our constant 8.314 Joules Per Mole Times Kelvin. We recall that again, temperature is in kelvin here and then our molar mass is in grams per mole. So we're going to begin by converting the given Speed of the private jet and mph to meters per second for our formula here. So we have 650 mph converting two m per second. So we're going to multiply first to cancel out our units of miles and we'll write mph as MPH in the denominator. So we're going to cancel out miles in the denominator by recalling that one mile has an equivalent of 1.6093 kilometers. So now canceling out miles, we want to now get rid of our unit kilometers by multiplying by our second conversion factor where one kilometer we would recall our prefix kilo tells us that it's equivalent to 10 to the third power of our base unit meters. Now canceling out kilometers. We're going to get rid of our unit hours and end up with seconds by multiplying by our last conversion factor where we recall that one hour in the numerator has an equivalent of 3600 seconds in the denominator. So now we can cancel out ours in the numerator and the denominator. And we're left with meters in the numerator and seconds in the denominator for our final units of meters per second. This is going to yield a speed of 290.568 m per second. So now that we have the speed, let's go into our root mean square speed formula. So we have Now 290. m/s plugged in for our root mean square speed or velocity. This is set equal to the square root of three, multiplied by r, gas constant. R which as we stated is 8.314 joules divided by moles times kelvin now multiplied by our temperature which is what we're solving for. We'll just plug it in as the variable T. Divided by our molar mass of our flooring molecules. So we're going to recall that from our periodic table. The molar mass of flooring gas would be the molar mass of one. Flooring atom times two. Since flooring gasses a diatonic molecule given by the subscript to here. So we would have a molar mass equal to g Permal for the flooring di atomic molecule. And we need to also recognize just to clarify further here that are units of molar mass. In our root mean square speed or velocity formula is equivalent to actually kilograms per mole. So we're going to convert from grams per mole to kilograms per mole. And We'll do so in the denominator here. So we have 38 g per mole. We're gonna multiply By placing g in the denominator so it can cancel out and kg in the numerator. And we would see that one g is equivalent to 10 to the negative third power kg. So this is going to simplify to 298.568 m/s equal to the square root of 24.9, jewels per mole times kelvin times temperature divided by 0. kg Permal. So we need temperature to be our final unit in kelvin and we want to cancel as many units as possible. So we need to get rid of this jewel unit here, we wanna make note of the conversion factor that one jewel is equal to one kg times meter square divided by seconds squared. So to simplify our next step further, we're going to get rid of that square root term by taking both sides to a power of two. And sorry, there should be an equal sign here. So on our left hand side we also have this race to an exponent of two. And in our next line we would simplify so that we have 844 to 9. m squared divided by second squared set equal to our right hand side where we no longer have that square root and we now have 24.942. We're going to interpret our jewels as now kilograms squared times meters squared divided by seconds squared, which is still divided by moles, times kelvin And multiplied in our numerator by our variable t. And now in our denominator, we want to plug in again that 0.03, eight kg Permal. So now focusing on getting rid of our units as well as simplifying further, we're going to cancel out our units of kilograms. And just to be clear, this should not be squared. It's just kilograms times meters squared times second squared equal to one jewel. So sorry for that mistake. We next want to cancel out our units of moles. And now we're going to simplify in our next line so that we have 844 to 9.76 m squared, divided by seconds squared equal to We would take 24. 42 divided by 0.38 in our calculators, which is going to yield a result of 656.3 meter square, divided by seconds squared. Still all divided by kelvin. Then multiplied by our variable temperature in kelvin. And now simplifying further, we're going to get rid of our numerator by dividing both sides by 656.3 68 m squared, divided by seconds squared so that it cancels out here. And then on our left hand side we have 656.3 m squared divided by seconds squared. Getting rid of those units now, Which will now simplify so that temperature is isolated and equivalent to 128.63. And we have our final unit being Kelvin here because this is all going to cancel out. So 128.63 Kelvin is our temperature. And now to get our final answer, we would say that our Kelvin temperature 128. Kelvin is equivalent to our degrees Celsius plus 273.15 and so two sulfur degree Celsius. We're going to subtract 2 73.15 from both sides Where we would finally get that our C temperature is equal to negative 144 0. degrees Celsius. And recalling from our prompt that we have three sig figs for the speed of the jet. We're going around this 23 sig figs As negative 145°C. And this would be our final answer as the temperature of our flooring gas molecules when they're moving at the same speed as the private jet at mph. So it's highlighted in yellow is our final answer. I hope everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.