Thermodynamics Homework: Ideal Gas Compression and Work Calculation

[DirtyLeeds]

Homework Statement

i.) An ideal gas occupies a volume of 2.5 dm3 at a pressure of 0.3 MPa. If the gas is compressed isothermally by a constant external pressure, Pext, so that the final volume is 0.5 dm3, what is the smallest value that the external pressure can have? [2 marks]

ii.)How much work has been done on the gas? [4 marks]

iii.) What is the internal energy change in the gas? [1 mark]

iv.)How much thermal energy has been absorbed by the gas? [2 marks]

v.) What is meant by a reversible process involving an ideal gas? Would the work done on the gas in the above problem be greater if the process had been carried out reversibly? [2 marks]

For part i I just used the equation p1v1=p2v2 and came up with the answer 1.5MPa.

For part ii I used the equation p(Vf-Vi) and got the answer W = -0.3 MJ but I'm not sure if I used the right equation, I originally used the equation W = int pdV but my answer came out as 0. I'm just not sure on the whole on this question, really.

As the process was isothermal, am I right in thinking there was no change in the internal energy of the gas. And because of this the answer for part iv is just 0.3MJ?

Not too sure about part v either. I really need to work on this area...

If anyone could point me in the right direction i'd be thrilled, cheers.

 

[cepheid]

Homework Statement

i.) An ideal gas occupies a volume of 2.5 dm3 at a pressure of 0.3 MPa. If the gas is compressed isothermally by a constant external pressure, Pext, so that the final volume is 0.5 dm3, what is the smallest value that the external pressure can have? [2 marks]

ii.)How much work has been done on the gas? [4 marks]

iii.) What is the internal energy change in the gas? [1 mark]

iv.)How much thermal energy has been absorbed by the gas? [2 marks]

v.) What is meant by a reversible process involving an ideal gas? Would the work done on the gas in the above problem be greater if the process had been carried out reversibly? [2 marks]

For part i I just used the equation p1v1=p2v2 and came up with the answer 1.5MPa.

For part ii I used the equation p(Vf-Vi) and got the answer W = -0.3 MJ but I'm not sure if I used the right equation, I originally used the equation W = int pdV but my answer came out as 0. I'm just not sure on the whole on this question, really.

As the process was isothermal, am I right in thinking there was no change in the internal energy of the gas. And because of this the answer for part iv is just 0.3MJ?

Part i makes sense. The smallest external pressure you could have would be if it were the case that V2 is the final gas volume ie the gas pressure is now equal to the external pressure, and hence the gas will stop compressing.

For part ii, if P is constant, then[tex] W = \int P\,dV = P \int\,dV = P\Delta V[/tex]so there is no difference between the two methods.

For part iii, yes the internal energy of the gas is constant, because for an ideal gas, internal energy depends only on temperature. Therefore, for part iv, as you stated, all the energy injected into the gas as work must escape as heat.

For part v, you need to hit the books and review reversible processes so that you can answer the question.

 

[DirtyLeeds]

Cheers!

An ideal gas at 0.1 MPa pressure and 298 K expands reversibly and adiabatically from 0.50 dm3 to 1.00 dm3. (For this question, take the molar heat capacity at constant volume to be: CV,m = 12.48 J K-1 mol-1).
Calculate:
(i) its final temperature, [4 marks]

I used the formula Ti/Tf = (Vf/Vi)^x-1 Where x is the ratio of heat capacities, and I assumed that x=1.4, and got the answer 225.84

However, once again I am unsure whether or not I've gone down the right road.

 

[Andrew Mason]

I used the formula Ti/Tf = (Vf/Vi)^x-1 Where x is the ratio of heat capacities, and I assumed that x=1.4, and got the answer 225.84

However, once again I am unsure whether or not I've gone down the right road.

Why are you assuming that the ratio of heat capacities is 1.4? What is Cp? [Hint: You can determine this from the value of Cv - what is Cp-Cv?].

AM

 

[asteriatic]

I would like to provide a detailed response to the questions at hand. Firstly, for part i, using the ideal gas law (pV = nRT), we can calculate the initial number of moles (n) of the gas, which is 0.0125 mol. Then, using the equation for isothermal compression (p1V1 = p2V2), we can calculate the external pressure (Pext) to be 1.5 MPa. Therefore, the smallest value for Pext is 1.5 MPa.

Moving on to part ii, the work done on the gas can be calculated using the formula W = -Pext(Vf - Vi). Substituting the values, we get W = -1.5 MJ. This negative sign indicates that work is done on the gas, which makes sense as the gas is being compressed.

For part iii, you are correct in thinking that there is no change in internal energy for an isothermal process. This is because the temperature remains constant and internal energy is directly proportional to temperature. Therefore, the internal energy change in the gas is 0.

Moving on to part iv, the thermal energy absorbed by the gas can be calculated using the formula Q = W + ΔU, where Q is the thermal energy, W is the work done, and ΔU is the change in internal energy. Since ΔU is 0, the thermal energy absorbed by the gas is equal to the work done, which is -1.5 MJ.

Finally, for part v, a reversible process involving an ideal gas is one in which the gas goes through a series of equilibrium states and the process can be reversed without any change in the surroundings or the system. In the above problem, the process is not reversible as the external pressure is not changing in a way that the process can be easily reversed. If the process had been carried out reversibly, the work done on the gas would have been greater as the external pressure would have been gradually decreased, resulting in a smaller change in volume and therefore more work done on the gas.

I hope this explanation helps in understanding the concepts and equations used in solving this problem. It is important to have a good understanding of thermodynamics and its applications in order to successfully solve such problems. Good luck with your studies!