Pizza Rolls 6 (pt.1)

[KDang_1D]

"... First you perform an isobaric compression of the container to 10.0L...."

Why is ∆U=0 for the first part of the problem if the process is isobaric, not isothermic?

Also, why is ∆S=0? Wouldn't a decrease in volume cause a decrease in number of possible positions?

[Hiba Alnajjar_2C]

In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without having to do calculations for each step of the process. Because the initial and final states are identical, the change in entropy and internal energy would be zero. In contrast, q and w are not state functions, so you have to calculate their specific values at each step and then add them all together to find the q and w for the entire process.

[Shail Avasthi 2C]

In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without having to do calculations for each step of the process. Because the initial and final states are identical, the change in entropy and internal energy would be zero. In contrast, q and w are not state functions, so you have to calculate their specific values at each step and then add them all together to find the q and w for the entire process.

Can you explain how you calculate q and w at each step of the process? I'm a bit lost.

[Hiba Alnajjar_2C]

In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without having to do calculations for each step of the process. Because the initial and final states are identical, the change in entropy and internal energy would be zero. In contrast, q and w are not state functions, so you have to calculate their specific values at each step and then add them all together to find the q and w for the entire process.

Can you explain how you calculate q and w at each step of the process? I'm a bit lost.

So, you can break up the problem into 3 steps. The first step is the isobaric compression from 100 L to 10.0 L. Because we know that delta U is equal to zero, we can state that q is equal to -w (or vice-versa). Therefore, you can set q equal to -P(delta V) since the pressure is constant. (Make sure to convert your quantity to Joules or kJ at every step). Once you find q, you can multiply it by -1 to find w. We know that w is going to be positive since this step compresses the system and work is being done on the system. For the second step, because the volume is constant, you know that w = 0. Since delta U is equal to 0 and equal to q+w, we can also conclude that q is also 0 for this step. For the final step, use the equation w = -nRT ln(V2/V1) to find the work. Once again, q will be the same quantity with the opposite sign. Then, add up all of the values found for q in each step to find the total q & do the same for w. Because q = -w, you can also just add up the values of q for each step and then change the sign of the total value of q to find the value of w (the totals should be equal and opposite). Hope this helps!

[Shail Avasthi 2C]

In this problem, delta S and delta U are equal to 0 when referring to the change in entropy/internal energy over the course of the entire process, which the question is asking for. Entropy and internal energy are state functions, meaning we can look at the initial and final states directly without having to do calculations for each step of the process. Because the initial and final states are identical, the change in entropy and internal energy would be zero. In contrast, q and w are not state functions, so you have to calculate their specific values at each step and then add them all together to find the q and w for the entire process.

Can you explain how you calculate q and w at each step of the process? I'm a bit lost.

So, you can break up the problem into 3 steps. The first step is the isobaric compression from 100 L to 10.0 L. Because we know that delta U is equal to zero, we can state that q is equal to -w (or vice-versa). Therefore, you can set q equal to -P(delta V) since the pressure is constant. (Make sure to convert your quantity to Joules or kJ at every step). Once you find q, you can multiply it by -1 to find w. We know that w is going to be positive since this step compresses the system and work is being done on the system. For the second step, because the volume is constant, you know that w = 0. Since delta U is equal to 0 and equal to q+w, we can also conclude that q is also 0 for this step. For the final step, use the equation w = -nRT ln(V2/V1) to find the work. Once again, q will be the same quantity with the opposite sign. Then, add up all of the values found for q in each step to find the total q & do the same for w. Because q = -w, you can also just add up the values of q for each step and then change the sign of the total value of q to find the value of w (the totals should be equal and opposite). Hope this helps!

That was really helpful. Thank you!

[Hailey Kim 4G]

Since internal energy (U) and entropy (S) are state functions, you can look at their initial and final values. Since they are the same (the system goes back to its original volume and pressure), the change of internal energy/entropy is 0.

[205150314]

What is T

[Leila_4G]

We are never given T so we can't solve the last step for w and therefore q.

[205150314]

We are never given T so we can't solve the last step for w and therefore q.

could you explain to me the problem please? like values

[Angela Prince 1J]

It is because you are considering delta U and delta S over the ENTIRE process, and because they are state functions, their pathway does not matter. In the problem, the final volume and pressure are the same so the change in either is 0.

[Angela Prince 1J]

We are never given T so we can't solve the last step for w and therefore q.

You can use pV=nRT to solve for T and then solve for work.