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

b. How much heat energy is transferred to the gas to cause this expansion?

Question

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

b. How much heat energy is transferred to the gas to cause this expansion?

Accepted Answer

Hi, everyone. In this practice problem, we are being asked to determine the amount of energy that must be added to a system to cause an expansion. We will consider a sealed container with a movable piston containing 0.1 moles of carbon dioxide gas and an initial pressure of 4:08 p.m. and an initial temperature of 300 Kelvin by maintaining a constant pressure, the carbon dioxide is expanded until its volumes is doubles. And we're being asked to determine the amount of energy that must be added to the system to actually cause this expansion. The options given are a 615 GB 230 jules, C 870 Joles and D 1000 100 and Joes. In this particular practice problem, it is given that the number of moles of carbon dioxide gas N will be equal to 0.1 mo And in this case, we want to assume that carbon dioxide is an ideal gas to make our calculations easier. So according to the ideal gas equation or ideal gas law, P, one multiplied by V one will be equals to N multiplied by R multiplied by T one. Therefore, we can actually rearrange our equation to first obtain our initial volume in this expansion. So that P one after rearrangement, we can obtain that by taking and multiply by R multiplied by T one divided by P one, we gonna calculate this in the first place. So that P one can then be obtained by N is going to be 0.1 more multiply that with the ideal gas constant or R which is going to be 8.314 jules divided by moles multiplied by Kelvin multiplied that with T one which is given in the problem statement to be 300 Kelvin or the initial temperature and divide all that with the uh initial pressure which will be for A PM which we have to convert into pascals to keep everything in um si unit. So four ATM multiplied by 1. multiplied by 10 to the power of five pascals divided by one ATM. So calculating all of this, we can obtain P one to then be equal to 0.615252 times 10 to the power of negative 3 m cube or in centimeter P one will then come out to be 615. centimeter cube P one just like so awesome. So now let's look at the isobaric expansion. It is given that P two or the final uh volume will double. So P two will be uh two times P one which uh we can just calculate that to be two multiplied by 615.25 centimeter cube, which will come out to be 1000 230. centimeter cube at constant pressure. The ideal guess relationship can be reduced into P two divided by T two equals to P one divided by T one. And what we are interested to find is DT two so that we can then calculate the total heat added to the system. So let's rearrange our equation so that T two can be calculated by taking P two multiplied by T one divided by P one. So let's substitute all of our values. So T two will then come out to BP two, which is going to be just two times P one. I'm just gonna keep it at that to make our calculation simpler, multiplied by T one. T one is given to be 300 Kelvin divide that with P one and P one is going to just remains as P one and we can just cross that out. So that T two will just be equal to 600 Kelvin. So heat added to the system can then be calculated or Q can be, can then be calculated by taking N multiplied by CP multiplied by delta T and N is going to be 0.1 mole. And the CP is going to be 37.4 Joles divided by Mole Kelvin and the delta T is going to be Kelvin minus 300. Kelvin, which is the difference between T two and T one. And that will give us the Q value of 1000 100 and 22 Jews, which will be the answer to this particular practice problem, which will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem. And I'll be it for this video. If you guys notice we actually do not need the value of P one and P two. And we can technically just uh take the relationship between P two and P one, which is equals to P two equals to two P one and put it into the isobaric expansion expression here to obtain the final T two to be 600 Kelvin and then substitute that value into the heat added to the system equation to obtain our option, the answer of 100 and 1000 100 and 22 Joel. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topic available on our website. But other than that, that'll be it. And thank you so much for watching.

5.0 g of nitrogen gas at 20°C and an initial pressure of 3.0 atm undergo an isobaric expansion until the volume has tripled. b. How much heat energy is transferred to the gas to cause this expansion?

Verified Solution

Key Concepts

Isobaric Process

First Law of Thermodynamics

Heat Capacity at Constant Pressure (Cp)