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

d. What is the reading of a pressure gauge attached to the tank?

Question

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

d. What is the reading of a pressure gauge attached to the tank?

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 to use in order to solve this problem. So a spherical container of diameter 240 millimeters is filled with 34 g of nitrogen gas at 35 degrees Celsius determine the pressure of nitrogen gas in the container as reported by a pressure gauge. OK. So we're given some multiple choice answers and they're all given in the same units of CALS. So let's read our multiple choice answers to see what our final answer might be. A is 2.9 multiplied by 10 to the fifth. Power B is 3.3 multiplied by 10 to the fifth. Power C is 1.9 multiplied by 10 to the fifth power and D is 4.3 multiplied by 10 to the fifth power. OK. So first off, let's assume that nitrogen is an ideal gas. So let's recall that the ideal gas equation is as small as we'll call it equation one. So the ideal gas law equation states that the pressure multiplied by the volume is equal to the number of moles multiplied by the universal gas constant multiplied by the temperature. Awesome. So then let's also recall the equation for pressure gauge is as fall and we'll call it is as falls and we'll call it equation two. I'm gonna denote it as pressure subscript G. So the pressure of the gauge is equal to the pressure minus the pressure of the atmosphere. P subscript atmosphere. Awesome. And also let's remember and recall a note that the volume of the gas, 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. Let's also make a quick note here that the universal gas constant, the numerical value for that is 8.314 and it's in units of jewels per mole multiplied by Calvin. And also let's note that one atmosphere is equal to 1. multiplied by 10 to the fifth power scales. Awesome. So our end goal is to find the pressure of nitrogen gas. So let's rearrange the ideal gas law, which in this case is I indicated it as equation one and solve for P, we'll call that equation three. So isolating pressure, solving for pressure alone, when we rewrite it, it will be pressure equals number of moles multiplied by the universal gas constant multiplied by temperature all divided by the volume. So in order to sulfur P first, we must determine what the, what the value for volume is. So let's do that really quick. So let's to do that. Let's recall what the volume of the sphere is. So the volume of sphere is V equals four thirds, four thirds multiplied by pi multiplied by radius. Qt OK. So let's plug in our known variables into the volume of it into the equation for the volume of the sphere determine to determine what the volume is. OK. So V equals four thirds multiplied by pi but we have to have a bit of a problem here. We'll have to take a little pause here to make a note here. We're given in our problem diameters instead of ra radius. So the diameter is 240 millimeters. So we need to remember and recall that radius equals diameter divided by two. So it's 240 millimeters divided by two and don't forget to the cube to the power of three. But then we have another problem is volume has to be in meters cubed in units of meters cubes. So we have to do a quick unit conversion. So using dimensional analysis, we should be able to recall and remember that in one m, there is 1000 millimeters and don't forget the cubed. So when you plug that into a calculator, your value for volume should be 7. multiplied by 10 to the minus three or I should say 10 to the power of negative three m cubed. Awesome. So we also need to determine the number of moles. So let's recall the equation to find the number of moles as the number of moles is equal to the mass divided by the molar mass. But we need to make a quick little side note here that nitrogen is diatomic. So there is two nitrogen atoms and a single molecule of nitrogen gas. So I'm gonna denote the molecular mass of nitrogen is capital m subscript, capital N subscript two. So the molecular mass of nitrogen gas, when you look it up or you determine it using a periodic table. So the molecular mass of nitrogen gas is 28.1 g per mole. So when we plug in all unknown variables into the mole equation, we're given 34 g of nitrogen gas and the molecular mass or the molar mass is 28.1 g per mole. The grams cancel out and are, and we come, when we, we plug into a calculator, we determine that the number of moles is 1.214 moles. OK. And then also at this stage, we're given the temperature as 35 degrees Celsius. So we need to convert it to Calvin. So let's recall how to do the unit conversion for that. It's really easy. So we just take our temperature degrees Celsius, which is 35 degrees and we add 273 which when we add those two together, we should get 308 Kelvin. OK. So when we plug in all of our known variables back into equation three, so we can finally solve for pressure, which was our end goal. So plugging in all of our known variables now, since we found all of our unknowns, so let's do that. The Solfur p. So the moles we found out was 1. mole multiplied by the universal gas constant, which was 8.314 Jews per mole multiplied by Calvin multiplied by a temperature which we found out in, in Calvin was 308 all divided by the volume which we determine the value for volume was seven point 238 multiplied by 10 to the power of minus three m cubed. So when you plug into a calculator, we should determine the pressure to be 4.294 multiplied by to the fifth power past scales. So then plugging in that value into equation to our pressure gauge equation. So plugging in our numerical values. So the pressure we found out is 4. multiplied by 10 to the fifth power past scales. And we might, we subtract it from the pressure of the atmosphere which we determined to be 1. multiplied by 10 to the fifth power pass scales. So when you put that into a calculator. The pressure of nitrogen gas as reported by the pressure gauge will be 3.3 times 10 to the fifth power has scales. All right, we found our final answer. Ok. So that means our final answer will be the letter B 3.3 multiplied by 10 of the fifth power pascals. Thank you so much for watching. Hopefully that helps 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. d. What is the reading of a pressure gauge attached to the tank?

Verified Solution

Key Concepts

Ideal Gas Law

Volume of a Cylinder

Pressure Measurement