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Work done by gas expansion on piston

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  • #1
CAF123
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Homework Statement


Consider a frictionless piston inside a cylinder, both with a cross sectional area A.

1) By considering the work done by the gas on the piston and the work dW required to move the piston a distance dx, show that the work done by the gas is dW = -pdV.
2) 1 L of an ideal gas is at a pressure of 5 atm. If it then expands at constant temperature against a frictionless piston until It's pressure falls back to atmospheric pressure, calculate I) the final volume of gas, 2) the work done on (or is it by?) the gas?

The Attempt at a Solution


1) The force of the gas on the piston is PA. Then W = Fdx = PA dx = PdV. This is straightforward but I have a positive work. Is there a mistake in the question? dx and F are in same direction so by definition, this is positive work.
2) I) 5L ii) So dV = 4 L . The work done will be by the gas since it is the gas that is expanding and the piston does not act to compress. I don't think I can use the above relation because the pressure is not constant. I will try an integral. Is this the right way to proceed?

EDIT: The integral I have is $$W = nRT\int_{V_o}^{V_f} \frac{dV}{V} = nRT[ln(V_f) - ln(V_o)] = nRT ln \left(\frac{V_f}{V_o}\right)$$ I would have then an answer of ##nRT ln(5)##, since I don't know n or T and R is a constant.
 
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Answers and Replies

  • #2
785
15

Homework Statement


Consider a frictionless piston inside a cylinder, both with a cross sectional area A.

1) By considering the work done by the gas on the piston and the work dW required to move the piston a distance dx, show that the work done by the gas is dW = -pdV.
2) 1 L of an ideal gas is at a pressure of 5 atm. If it then expands at constant temperature against a frictionless piston until It's pressure falls back to atmospheric pressure, calculate I) the final volume of gas, 2) the work done on (or is it by?) the gas?

The Attempt at a Solution


1) The force of the gas on the piston is PA. Then W = Fdx = PA dx = PdV. This is straightforward but I have a positive work. Is there a mistake in the question? dx and F are in same direction so by definition, this is positive work.
First question :

Is the pressure "P" , external pressure or pressure of the gas ? Also is this question from thermodynamics , or chemical energetics ? I am asking because , work done by/on gas in chemistry is written -Pdv , while in physics , its written as PdV , depending on the sign of dV.

2) I) 5L ii) So dV = 4 L . The work done will be by the gas since it is the gas that is expanding and the piston does not act to compress. I don't think I can use the above relation because the pressure is not constant. I will try an integral. Is this the right way to proceed?

EDIT: The integral I have is $$W = nRT\int_{V_o}^{V_f} \frac{dV}{V} = nRT[ln(V_f) - ln(V_o)] = nRT ln \left(\frac{V_f}{V_o}\right)$$ I would have then an answer of ##nRT ln(5)##, since I don't know n or T and R is a constant.
Second question :

So far , so good. Now use PV=nRT and Boyle's law to eliminate "T" and write the given expression in terms of initial and final pressures and initial volume. Also , these two things can help you find final volume also.

Edit: I assume this is a physics question. Thus work done on gas is negative. If it is from chemistry, then work done on gas will be positive. Vice versa for work done by gas....
 
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  • #3
CAF123
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First question :

Is the pressure "P" , external pressure or pressure of the gas ? Also is this question from thermodynamics , or chemical energetics ? I am asking because , work done by/on gas in chemistry is written -Pdv , while in physics , its written as PdV , depending on the sign of dV.
It is pressure from the gas and the question is from a physics course on thermodynamics. Would you say there is an error in the question regarding that minus sign? I see your edit as I previewed my post - so there is an error just to clarify? Why the different convention across the two branches?

Second question :

So far , so good. Now use PV=nRT and Boyle's law to eliminate "T" and write the given expression in terms of initial and final pressures and initial volume.
Is ##P_i V_i = P_f V_f## what you mean? How can I use this to substitute for T?
 
  • #4
785
15
It is from a physics course on thermodynamics. Would you say there is an error in the question regarding that minus sign? I see your edit as I previewed my post - so there is an error just to clarify? Why the different convention across the two branches?
That was written in my textbook. Well ,in chemistry we are more worried about "internal energy" changes in a body. If you do work on the gas (consider gas as the system), then you will have to sacrifice your energy and the internal energy of the gas will increase. Hence work done on the gas is positive. In physics ,we think in a mechanical way. You do work on the gas means the force of yours on the gas is trying to resist the motion of gas to move the piston. Displacement opposite to force. Hence work done on gas is negative. Again, in your first question, is "P" external pressure or pressure of the gas itself ?

Is ##P_i V_i = P_f V_f## what you mean? How can I use this to substitute for T?
PiVi=PfVf=nRT , by combining Boyle's law and ideal gas equation. Now you can use this to get your way out.

Also , while you calculate the value for work done on the gas, the question being from physics, you should get work done on gas negative in this case.
 
  • #5
CAF123
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Again, in your first question, is "P" external pressure or pressure of the gas itself ?
I just spoke to my professor and he said both interpretations are valid (i.e you could say positive work is done by the gas or negative work is done by the gas). So there is not an error in the question - it is just a matter of what exactly you are considering. P is the pressure by the gas.


Using your suggestion, I have $$nR \left(\frac{P_f V_f}{nR}\right) \left[ln\frac{V_f}{V_i} \right] = 5 ln (5) .$$
 
  • #6
785
15
I just spoke to my professor and he said both interpretations are valid (i.e you could say positive work is done by the gas or negative work is done by the gas). So there is not an error in the question - it is just a matter of what exactly you are considering. P is the pressure by the gas.
Well -PdV as the answer is correct if pressure is external in case of expansion of gas. When did I said that "always" and that "there is error in question"? In physics, we think mechanically. Also I assume external pressure. Then only force and displacements are opposite to each other. See, dV is always positive and sign of expression is depending on "P". If P is pressure exerted "by" gas, how can work done be negative by gas ?

Using your suggestion, I have $$nR \left(\frac{P_f V_f}{nR}\right) \left[ln\frac{V_f}{V_i} \right] = 5 ln (5) .$$
Will you not put "minus" sign as you calculate work done on the gas, during expansion ?
 
  • #7
CAF123
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Well -PdV as the answer is correct if pressure is external in case of expansion of gas. When did I said that "always" and that "there is error in question"? In physics, we think mechanically. Also I assume external pressure. Then only force and displacements are opposite to each other.
Why are F and dx in opposite direction?
 
  • #8
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With all due respect to your professor, your understanding is correct and his is wrong. The work done by the gas on the piston (and adjacent surroundings) is not -pdV, it is + pdV. And the work done by the surroundings (and the piston) on the gas is -pdV. So the force and displacement for the gas on the surroundings are in the same direction, and the force and displacement for the surroundings on the gas are in opposite directions (when you consider that dV for the gas is -dV for the surroundings).

Incidentially, the work you calculated applies only if the piston moves very slowly, so that all the gas in the cylinder is at a uniform pressure. It is also possible to apply a sudden drop in the pressure imposed by the surroundings on the piston, in which case the pressure within the cylinder immediately adjacent to the piston will match the pressure of the surroundings at all times (for a massless piston), but the gas pressures away from the piston face will not. To calculate the work on the surroundings in this case, you will not have convenient access to the pressure within the cylinder at the piston face, and it will be much more convenient to impose and measure the force per unit area exerted by the surroundings on the opposite face of the piston. This is called the "external pressure," pex. Since, for the case of a massless piston, the pressure exerted by the gas inside the cylinder on the piston face must match the external pressure imposed by the surroundings on the opposite piston face, the external pressure can be used to calculate the work done by the gas on the surroundings. That is, dW = pexdV.

I would also like to point out that, in your equation 5 ln(5), it is only a coincidence that both numbers are 5. The first 5 has dimensions of work (liter-atm), but the second 5, representing the ratio of the volumes, is dimensionless. In other problems, the numbers in the two locations will not match.
 
  • #9
785
15
Why are F and dx in opposite direction?
Imagine,imagine and Imagine !! Gas is expanding due right suppose, then will not external pressure pose hindrance to left ?


With all due respect to your professor, your understanding is correct and his is wrong. The work done by the gas on the piston (and adjacent surroundings) is not -pdV, it is + pdV. And the work done by the surroundings (and the piston) on the gas is -pdV. So the force and displacement for the gas on the surroundings are in the same direction, and the force and displacement for the surroundings on the gas are in opposite directions (when you consider that dV for the gas is -dV for the surroundings).
That was what I was trying to explain OP for so long. :smile:

Incidentially, the work you calculated applies only if the piston moves very slowly, so that all the gas in the cylinder is at a uniform pressure.
Yes, reversible process...

It is also possible to apply a sudden drop in the pressure imposed by the surroundings on the piston, in which case the pressure within the cylinder immediately adjacent to the piston will match the pressure of the surroundings at all times (for a massless piston), but the gas pressures away from the piston face will not. To calculate the work on the surroundings in this case, you will not have convenient access to the pressure within the cylinder at the piston face, and it will be much more convenient to impose and measure the force per unit area exerted by the surroundings on the opposite face of the piston. This is called the "external pressure," pex. Since, for the case of a massless piston, the pressure exerted by the gas inside the cylinder on the piston face must match the external pressure imposed by the surroundings on the opposite piston face, the external pressure can be used to calculate the work done by the gas on the surroundings. That is, dW = pexdV.
Yes, dW = pexdV is the work done by gas against external pressure and work done by external pressure on gas in his first question will be -PexdV.

I would also like to point out that, in your equation 5 ln(5), it is only a coincidence that both numbers are 5. The first 5 has dimensions of work (liter-atm), but the second 5, representing the ratio of the volumes, is dimensionless. In other problems, the numbers in the two locations will not match.
Yes , but since he is finding work done on gas, he should put a "minus" sign in his answer : i.e. -5ln(5)...
 
  • #10
CAF123
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Imagine,imagine and Imagine !! Gas is expanding due right suppose, then will not external pressure pose hindrance to left ?
Say I have the gas expanding to the right. So it expands and thus the piston moves a little dx. If F is the force of the gas that moves the piston , then F is in the same direction as dx,no? If F and dx were in opposite directions, then the work done by the gas would be negative?
EDIT: I see your talking about external pressure. This makes sense now - external pressure like, for example, the piston compressing the gas?
 
  • #11
785
15
Say I have the gas expanding to the right. So it expands and thus the piston moves a little dx. If F is the force of the gas that moves the piston , then F is in the same direction as dx,no? If F and dx were in opposite directions, then the work done by the gas would be negative?
EDIT: I see your talking about external pressure. This makes sense now - external pressure like, for example, the piston compressing the gas?
Yes, you have a gas 1 at one side of a partitioned box and gas 2 at the other side. For gas 1 pressure on other side will be external pressure and vice versa for gas 2. But you are correct.
 

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