Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20°C undergo an isobaric expansion until the volume has tripled.

a. What is the gas volume after the expansion?

Question

Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20°C undergo an isobaric expansion until the volume has tripled.

a. What is the gas volume after the expansion?

Accepted Answer

Hey, everyone. Let's go through this practice problem. A container is filled with 10 g of helium gas at a temperature of degrees Celsius and a pressure of 2.5 atmospheres. The volume of the gas doubles due to isobaric expansion, determine the final volume of the gas. We're given four multiple choice options to choose from. Option A 24,446 cubic centimeters. Option B 48,916 cubic centimeters, option C 48,892 cubic centimeters and option D 24,458 cubic centimeters. Now, we're looking for the gas's final volume and the problem explicitly tells us that throughout the ex over the course of the expansion, the volume of the gas doubles. So the final volume which I'm going to write as V sub two is equal to twice the initial volume, which I'm going to write as V sub one. So I'm going to use the one subscript to refer to initial variables and the two subscripts to refer to ending variables. But what this relation tells us is that to find the final volume, we first just need to find the initial volume. And we're given a bunch of other variables about the gasses, initial state from its mass, its initial temperature and its initial pressure. So we can find a relationship between these variables to find the initial volume by using the ideal gas law. And what the ideal gas law states is that the pressure of a gas multiplied by its volume is equal to the number of moles multiplied by the ideal gas constant R multiplied by the temperature. Since we're looking for V sub one, I'm going to add the one subscript next to the appropriate variables. And since we want an equation for V sub one, we just algebraically solve this for V sub one by dividing both sides of the equation by P sub one. So what we find is that the initial volume is equal to N RT divided by the initial pressure. So in order to use this equation, we first need to establish the other variables given to us in order to put them into the equation. So of course, first recall that the ideal gas constant R has a value of 8.314 Jews per mole. Kelvins. And the initial temperature T sub one is given to us as 25 degrees Celsius. But we need to convert this into Kelvins to use it in the problem in the ideal gas law. So we have to add 273 to it And so the initial temperature is equal to 298 Kelvins. We also need to know the initial pressure that's given to us in the problem as 2.5 atmospheres. But in order to use it in the ideal gas law, we need to convert it into pascals. So the conversion rate for that is that there are 1.13 multiplied by 10 to the power of five pascals for one atmosphere. And putting that in, we find an initial pressure of about 253, pascals. So that's our initial pressure. The final thing we need is N the number of moles of the gas. And this part can be kind of tricky because we're told the mass of the gas we have. But we need to convert that into a number of moles. And we can find that using a molar mass, we're told that the gas is helium. And we know that the molar mass of helium is four g per mole. So we can use the molar mass of helium as a conversion factor to convert between the mass of the helium to its number of moles. So let's try it. So we're told that there are 10 g of the helium gas and then using the molar mass. So that is one mole for four g. And if we, so basically, we're just dividing 10 by four, which gives us a number of moles of 2.5. And now we have everything we need. So let's put all this into the V sub two equation or the, the V sub one equation to solve for the initial volume. So that's the number of moles, 2.5 moles multiplied by the ideal gas constant, 8. jewels per mole album multiplied by the temperature. 298 Kelvins all divided by the initial pressure, 253,000 250 pascals. And if you put this into a calculator, then we find an initial volume of about 0. 58 cubic meters. And as we discussed earlier, the final volume could be twice the initial volume. So we're going to multiply this by two. So V sub two is equal to two multiplied by V sub one. So if we just take the value we just calculated for and multiply it by two, then if we find a final volume of about 0. 916 cubic meters. However, if we look at the multiple choice options, we're given all the options we have are in terms of cubic centimeters. So we're going to have to do a conversion from cubic meters to cubic centimeters. So of course, there are 100 centimeters in one m. So we use this as our conversion factor. And so we find a final answer in cubic centimeters of about 48, 916 cubic centimeters. And so with that, then is the answer to this problem. And if we look at our options, we can see that this agrees with option B which was 48,916 cubic centimeters. So option B is our answer to this problem and that's it for this video. I hope this video helped you out. If you think you need more practice, please check out some of our other tutoring vis videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

Five grams of nitrogen gas at an initial pressure of 3.0 atm and at 20°C undergo an isobaric expansion until the volume has tripled. a. What is the gas volume after the expansion?

Verified Solution

Key Concepts

Ideal Gas Law

Isobaric Process

Charles's Law