What is the Work Done by a Gas in a Cycle?

[knc]

Homework Statement

Homework Equations

PV = nRT \\ W = - \int_{V_i}^{V_f} P dV
\Delta E_{int} = Q + W

The Attempt at a Solution

a)[/B]
Since this is a cyclic process, the change in internal energy of the system is 0.
\Delta E_{int} = 0
The process causes some ice to melt, meaning heat transfers out of the system.
To maintain the model of a cyclic process the work being done on the gas (positive) is equal-but-opposite to the heat transferring out of the gas (negative)
Heat transferring into the water:
Q = m L_f \\<br /> W = - (-Q) = m L_f
This is intuitive and makes sense, however I don't understand the relevance of the very quickly part.
I do understand that this suggests this is not a quasi-static process and that the system is not at equilibrium at all times. but I don't see what the implications of that are.

b)
Rearranging the ideal gas law and plugging into work equation:
W = - \int_{4v_2}^{v_2} \frac{nrT}{V} dV
I don't think I am doing this part correctly.

c)
I don't know where to begin with this.

 

[Chestermiller]

Homework Statement

View attachment 200000

Homework Equations

PV = nRT \\ W = - \int_{V_i}^{V_f} P dV
\Delta E_{int} = Q + W

The Attempt at a Solution

a)[/B]
Since this is a cyclic process, the change in internal energy of the system is 0.
\Delta E_{int} = 0
The process causes some ice to melt, meaning heat transfers out of the system.
To maintain the model of a cyclic process the work being done on the gas (positive) is equal-but-opposite to the heat transferring out of the gas (negative)
Heat transferring into the water:
Q = m L_f \\<br /> W = - (-Q) = m L_f

The heat is transferred out of the water to melt the ice. So, Q is negative. The work done on gas is -Q, which is a positive number. So your answer for the net work done on the gas is correct.

This is intuitive and makes sense, however I don't understand the relevance of the very quickly part.
I do understand that this suggests this is not a quasi-static process and that the system is not at equilibrium at all times. but I don't see what the implications of that are.

The implication is just that the gas suffers an unspecified irreversible compression during this step. Since the total heat removed during the cycle is specified in the problem statement, we don't need to know the details of this step.

b)
Rearranging the ideal gas law and plugging into work equation:
W = - \int_{4v_2}^{v_2} \frac{nrT}{V} dV
I don't think I am doing this part correctly.

You have the right idea, but the integration limits are flipped.

c)
I don't know where to begin with this.

Hint: Since, in step A, the piston is moved very quickly, there is no time for heat transfer in this step. You know the heat transferred for the entire cycle, and you know the heat transferred in step C.

 

[knc]

You have the right idea, but the integration limits are flipped.

In step A the piston compressing the gas; it's going from its original volume (4V_2) to its compressed volume (V_2).
Even then, are we assuming that the gas is compressed so quickly that the temperature remains constant?
That doesn't seem right.

Also, the hint for the problem in part B threw me off. Why do we have to "Think about the other steps" if we are just using the work integral?

 

[Chestermiller]

In step A the piston compressing the gas; it's going from its original volume (4V_2) to its compressed volume (V_2).
Even then, are we assuming that the gas is compressed so quickly that the temperature remains constant?
That doesn't seem right.

We're not saying that the temperature remains constant. We are saying that no heat is transferred between the gas and the ice water. Do you know why this is not the same as saying that the temperature remaining constant?

 

[knc]

We're not saying that the temperature remains constant. We are saying that no heat is transferred between the gas and the ice water. Do you know why this is not the same as saying that the temperature remaining constant?

I see.
Although the temperature is not constant, the process is done so quickly as to not transfer heat?

 

[Chestermiller]

I see.
Although the temperature is not constant, the process is done so quickly as to not transfer heat?

Yes! Exactly! (Now, don't you like this site better than Physics Stack Exchange?)

 

[knc]

Of course I like this site better but when it's Sunday afternoon I cannot leave any stone unturned. :)

We have determined in step A that there is no heat flow out of the system. But we still cannot use the integral to solve for work?

 

[Chestermiller]

Of course I like this site better but when it's Sunday afternoon I cannot leave any stone unturned. :)

We have determined in step A that there is no heat flow out of the system. But we still cannot use the integral to solve for work?

We don't need to know the work in step A if we know that the heat is zero. We can integrate to get the work in step C, and this will give us the heat in step C. So the heat in step B is the total heat minus the heat in step C.

 

[knc]

In step C, since the piston is being raised slowly, we can say that the temperature is always in equilibrium, correct?
So the work done on the gas in step C:
W = - \int_{V_2}^{4V_2} \frac{nRT}{V}dV \\ W = - nRT \ln{4}
How can we relate this to the heat transfer in step C?

I understand your reasoning, I think:
Since step A is done very quickly, it's adiabatic.
So the only heat transfer being done are in steps B and C.
Since we have an expression for total heat transfer, we can solve for one and have the other.

 

[Chestermiller]

In step C, since the piston is being raised slowly, we can say that the temperature is always in equilibrium, correct?
So the work done on the gas in step C:
W = - \int_{V_2}^{4V_2} \frac{nRT}{V}dV \\ W = - nRT \ln{4}
How can we relate this to the heat transfer in step C?

Step C is done very slowly, so it is isothermal at 0 C. That means that $\Delta E=0$ for this step.

 

[knc]

Step C is done very slowly, so it is isothermal at 0 C. That means that $\Delta E=0$ for this step.

Sorry for the late response...

step C:
$\Delta E_{int} = 0 \\ W = -Q \\ \therefore Q = nRT \ln{4}$

This is our heat transfer in step C.

Heat in step B:
$Q = mL_f - nRT \ln{4}$

Does this seem right?

 

[knc]

Could we apply this same model to solve for work done in step A, seeing as there is only work done in steps A and C?

 

[Chestermiller]

Sorry for the late response...

step C:
$\Delta E_{int} = 0 \\ W = -Q \\ \therefore Q = nRT \ln{4}$

This is our heat transfer in step C.

Heat in step B:
$Q = mL_f - nRT \ln{4}$

Does this seem right?

The overall heat flow for the cycle is negative (-mL), so the heat flow in step B is $Q =- mL_f - nRT \ln{4}$.

 

[Chestermiller]

Could we apply this same model to solve for work done in step A, seeing as there is only work done in steps A and C?

Yes. Let's see what you get.

 

[knc]

The overall heat flow for the cycle is negative (-mL), so the heat flow in step B is $Q =- mL_f - nRT \ln{4}$.

So is the question misworded? By convention, I thought that a negative heat represented heat leaving a system whereas positive heat represented heat flowing into a system?

Edit: Nevermind, it makes sense. The energy is transferring out of the gas so we denote the heat flow to be negative.

Although the negative sign is intuitive; the gas is warmer than the surrounding ice-water mixture and heat flows out until the system reaches equilibrium.

 

[knc]

Yes. Let's see what you get.

$W_a + W_c = W_{net} \\ W_a = m L_f - W_c \\ W_c = -nRT \ln{4} \\ W_a = m L_f + nRT \ln{4}$

 

[knc]

Does this mean the work done in step A is equal-but-opposite to the heat transfer in step B? Does this make sense?

 

[Chestermiller]

$$W_{total}=-Q_{total}=+mL_f$$
$$W_C=nRT\ln{4}$$
$$W_A=W_{total}-W_C=mL_f-nRT\ln{4}$$

 

[knc]

$$W_{total}=-Q_{total}=+mL_f$$
$$W_C=nRT\ln{4}$$
$$W_A=W_{total}-W_C=mL_f-nRT\ln{4}$$

So in step C, $\Delta E_{int} = 0$
$$ Q_c + W_c = 0 \\ Q_c = - W_c $$
Using integral to solve for W_c:
$$ W_c = - \int_{V_2}^{4V_2} \frac{nRT}{V} dV \\ W_c = - nRT \ln{4} $$

How did you get positive work done? The piston is being raised, so isn't the gas doing work on its surroundings? Hence the negative?

 

[Chestermiller]

So in step C, $\Delta E_{int} = 0$
$$ Q_c + W_c = 0 \\ Q_c = - W_c $$
Using integral to solve for W_c:
$$ W_c = - \int_{V_2}^{4V_2} \frac{nRT}{V} dV \\ W_c = - nRT \ln{4} $$

How did you get positive work done? The piston is being raised, so isn't the gas doing work on its surroundings? Hence the negative?

Oops. Sorry. You're right. My bad.

 

[knc]

Oops. Sorry. You're right. My bad.

So is it correct that the heat transfer in step B is equal to opposite of work done in step A?
$$ Q_b = - W_a $$

 

[Chestermiller]

So is it correct that the heat transfer in step B is equal to opposite of work done in step A?
$$ Q_b = - W_a $$

Sure. Now, can you explain why?

 

[knc]

$$ Q_{net} = - W_{net} $$
$$ Q_a + Q_b + Q_c = -W_a - W_b - W_c $$
$$ \Delta E_{int,c} = \Delta E_{int,a} = 0 $$
$$ Q_a = -W_a \\ Q_c = - W_c $$
$$ -W_a + Q_b - W_c = -W_a - W_b - W_c $$
$$ Q_b = -W_a $$
Mathematically it makes sense. What physical implications does this hold?

 

[Chestermiller]

Step B brings the temperature back down to the value before Step A. So there is no change in internal energy over steps A and B. Work is done in step A but no heat; heat is done in step B, but no work.

 

[knc]

Thanks for the help.