A cylinder contains gas at a pressure of 2.0 atm and a number density of 4.2 x 10²⁵ m⁻³. The rms speed of the atoms is 660 m/s. Identify the gas.

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Hey, everyone. So this problem is dealing with the root means square speed of gasses. Let's see what it's asking us. We have a sealed container that has a pressure of 1. atmospheres and contains gas molecules with a root mean squared speed of 343 m per second at a temperature of 300. Kelvin. We're asked to name the gas in the container. If it has a number density of 5.8 times 10 to the 25th meters to the negative three. Our multiple choice answers are a helium B Krypton C neon or D argo. OK. So the first thing we need to do to solve this problem is recall our equation for the um root means squared speed with respect to pressure. And so when we're dealing with gasses, we can recall the equation PV equals one third and which is our number of molecules in the um container multiplied by Mr mass multiplied by our speed. This is our regular speed uh B to the negative two. And then we can also recall that the root means square speed is equal to the square root of our A V to the negative two. And so we can plug that in and our equation becomes P our, sorry, our pressure multiplied by VR volume. That's a capital V is equal to one third and multiplied by M multiplied by VR MS. And that's squared. And so we're told from the problem that our root mean squared speed is equal to 343 m per second. Our pressure is 1. atmospheres and then we have a number density. So our number density is going to be our total number of molecules divided by our volume. And that is equal to 5.8 times 10 to the meters to the negative three, we are asked to bind the gas in the container and we can use the mass to identify the atomic number. And then we can use that to identify the gas. So we'll be solving four mass in this problem. So we rearrange this so that we have mass on the left hand side of the equation isolated. And so we'll have mass is equal to three P B divided by VR MS squared multiplied by N. And so when we go to plug in those values, we have M is equal to three. Now, our pressure was 1.5 atmospheres. We need to keep everything in standard units. So we can recall our conversion factor between pascals and atmospheres is 101, pascals per atmosphere. Then we went to divide all of that by our routine squared speed. So 343 m per second, that value is squared. And then this problem has a volume divided by our number of molecules because we were given number of the number density which is an over B that goes in the denominator. And so we're gonna multiply that by 5.8 times 10 to the 25th meters, the negative three, we plug that into our calculator and we get 6.68 times 10 to the negative 26 kg. And then so from there, we can use our conversion factor between atomic mass units and a unit of U and kilograms. So we'll have 6.68 times 10, the negative 26 kg multiplied by one U divided by 1.66 text 10 to the negative 27 kg. And that gives us 40 view and we can look at our periodic table and recognize that that is the atomic mass for ARGO. And so when we look at our multiple choice answers that aligns with answer choice D so D is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

A cylinder contains gas at a pressure of 2.0 atm and a number density of 4.2 x 10²⁵ m⁻³. The rms speed of the atoms is 660 m/s. Identify the gas.

Verified Solution

Key Concepts

Ideal Gas Law

Root Mean Square (RMS) Speed

Number Density