Isotherm, Isobaric, and Isochoric Problem

[physics4u]

Homework Statement

One mole of a monatomic ideal gas has an initial pressure, volume, and temperature of Po, Vo, and 442 K, respectively. It undergoes an isothermal expansion that triples the volume of the gas. Then the gas undergoes an isobaric compression back to its original volume. Finally the gas undergoes an isochoric increase in pressure, so that the final pressure, volume and temperature are Po, Vo, and 442 K, respectively.

Find the total heat for this 3-step process?

Homework Equations

W=nRTln(Vf/Vi)

The Attempt at a Solution

For the first part of this question I used w=nRTln(Vf/Vi) and that's how I would find my Q(heat) for the isotherm b/c Q=W since temp is constant. But since Volume is not given to us, and it says the volume triples I just put in the value for 3 in volume and i ended up with 4037J

The second part of the problem for Isobaric I don't know how to solve. Because the eqn I was planning on using is W=P(Vf-Vi) but since there was no volume given I wasn't able to solve for this problem. So how do I solve for Q for the Isobaric?

For the third part, the Isochoric, I just used 3/2nRT and I ended up with 5512.2J.

I know I should be adding all the Q's(Q1+Q2+Q3) to get my total heat but how do I solve for Heat for the Isobaric part, when I am not given volume or pressure?

 

[anathema]

Thank you for your question. Let's break down the problem into the three steps and solve for the heat in each step.

First, for the isothermal expansion, you correctly used the equation Q=W=nRTln(Vf/Vi). Since you know the initial volume and temperature, you can solve for the final volume by using the fact that the volume triples. This means that the final volume will be 3 times the initial volume, or 3Vo. Plugging this into the equation, we get Q=nRTln(3). Using the ideal gas law, we can solve for nRT which is equal to PoVo. Substituting this in, we get Q=PoVo ln(3). Plugging in the values given, we get Q=4037 J.

For the isobaric compression, we can use the equation W=P(Vf-Vi). Since the final volume is equal to the initial volume, we can cancel out the Vf and Vi terms and we are left with W=P(Vo-Vo)=0. This means that no work is done during the isobaric compression, so the heat will be equal to 0.

Finally, for the isochoric increase in pressure, we can use the equation Q=nCv(Tf-Ti), where Cv is the molar specific heat at constant volume. Since the temperature is constant, the change in temperature will be 0, and therefore the heat will also be 0.

To find the total heat for the 3-step process, we simply add up the heat from each step. So the total heat for this process will be Q=4037+0+0=4037 J.

I hope this helps and good luck with your studies!

 

[Moetasim]

I would approach this problem by first understanding the definitions of isotherm, isobaric, and isochoric processes. An isotherm process is one in which the temperature remains constant, while an isobaric process is one in which the pressure remains constant, and an isochoric process is one in which the volume remains constant.

In this problem, we are given that the initial temperature is 442 K and the initial pressure is Po. We also know that the gas undergoes an isothermal expansion, which means that the temperature remains constant throughout this process. Using the equation Q=W=nRTln(Vf/Vi), we can calculate the heat for this process by plugging in the given values for n (1 mole), R (the ideal gas constant), and the initial and final volumes. This gives us a value of 4037 J.

For the second part of the problem, we are asked to find the heat for the isobaric compression back to its original volume. Since the volume is not given, we can assume that it remains constant, as it is stated that the final volume is the same as the initial volume. Therefore, we can use the equation Q=W=PΔV, where P is the constant pressure and ΔV is the change in volume (which is 0 in this case). Thus, the heat for this step is 0 J.

Finally, for the third part of the problem, we are asked to find the heat for the isochoric increase in pressure, which means the volume remains constant. In this case, we can use the equation Q=ΔU, where ΔU is the change in internal energy. Since the volume is constant, the change in internal energy is equal to the change in temperature, which in this case is 0 K. Therefore, the heat for this step is also 0 J.

To find the total heat for the 3-step process, we simply need to add up the heat values for each step: 4037 + 0 + 0 = 4037 J. This is the total heat for the entire process. It is important to note that the heat values for the isobaric and isochoric steps are 0 because no work is done in these processes, as the volume and pressure remain constant.

In conclusion, as a scientist, I would approach this problem by understanding the concepts of isotherm, isobaric