A 20-cm-diameter cylinder that is 40 cm long contains 50 g of oxygen gas at 20°C.

c. What is the number density of the oxygen?

Question

A 20-cm-diameter cylinder that is 40 cm long contains 50 g of oxygen gas at 20°C.

c. What is the number density of the oxygen?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need in order to solve this problem. A sphere of diameter 220 millimeters holds 36 g of chlorine gas at room temperature. 25 degrees Celsius calculates the number density of chlorine gas in the container. Awesome. So we're given some multiple choice answers and they're all in the units of molecules per meters cubed. So let's read it off, read them all off and see what our final answer might be. So A is 1.2 multiplied by 10 to the 26th. Power B is 5.5 multiplied by 10 to the 19th. Power C is 5.5 multiplied by 10 to the 25th. Power D is 5.5 multiplied by 10 to the 22nd power and E is 1.2 multiplied by 10 to the 20th power. Awesome. So to solve for the number density, we must recall that the number density equation is as false. We're gonna write the number density as capital N capital D is equal to the number of particles capital N divided by capital V the volume. So also recall that the volume of the gas, which I'm gonna write it as V subscript G is equal to the volume of the container which I'm gonna write as V subscript C. So to solve for the volume, we must recall and use the volume of a sphere contain the volume of a sphere equation. So the volume of the sphere is equal to. So V of a. So for the volume sphere, so V equals four thirds multiplied by pi multiplied by radius qed. So let's plug in our known variables and solve for the numerical value for the volume. So V equals four thirds multiplied by pi multiplied by the rays. But we need to note that our in our initial problem, we're given it as the diameter. So the radius, so radius equals diameter divided by two. OK. So let's plug that in. So 220 millimeters divided by two, don't forget the cub, but then we have to convert millimeters to meters. So let's do a quick unit conversion. So in one m there is 1000 millimeters and don't forget the cube to keep things consistent. So when you plug that into a calculator, the numerical value you should get is 5.575 multiplied by 10 to the negative three m scute. OK. So we need to recall that the number of particles the equation for the number of particles is capital N is equal to the number of moles lowercase N multiplied by capital N subscript capital A Avogadro's number. So the number of molecules or the number of particles, I should say the number of particles is equal to the number of moles multiplied by Avogadro's number. Awesome. Also, let's quickly recall that the number of moles is equal to the mass divided by capital and the molar mass. OK. So considering we're gonna call the number of particles equation number one. Equation number one, and we're gonna call the number of moles equation equation two. So considering equation two, we can rewrite equation one as following. So we can plug in the number of moles, we could plug in little N which was we disco discovered or remember the mass divided by the molar mass, we could plug that in for N. So we can, so we can get equation three. So the number of particles is equal to the mass divided by the molar mass multiplied by Avogadro's number. So also please note that when you look up the molar mass of chlorine note that there is two chlorine molecules because it's diatomic. So considering that and let's make a note here. So the molar mass of chlorine, which I'm gonna denote it as capital M subscript C L subscript two. So the molecular mass of chlorine is 70.9 g per mole. So you want, you can quickly determine that yourself or you can look it up. So when we plug in the numerical values into equation three, we're given 36 g of the sample. And the molar mass, as we just said is 70.9 g per mole multiplied by Avogadro's number, which the numerical value for Ava God's number is 6.22 multiplied by to the 23rd power. And it's in units of molecules Perel. So when you plug that into a calculator, you should get 3.577 multiplied by 10 to the 23rd molecules. OK. So finally to solve for our final answer. So number density, when we plug in our numerical values, we plug in our number of particles which we found out was 3.577 multiplied by 10 to the 23rd molecules. And we divide it by the volume which we determined was 5.575 multiplied by 10 to the negative third power meters cube. So when you plug that all into a calculator, you should get 5.5 times 10 to the 25th molecules, four m skewed. And that is our final answer. Yeah. So that means our final answer is C 5.5 multiplied by 10 to the 25th power. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

A 20-cm-diameter cylinder that is 40 cm long contains 50 g of oxygen gas at 20°C. c. What is the number density of the oxygen?

Verified Solution

Key Concepts

Number Density

Ideal Gas Law

Volume of a Cylinder