# Work done by an expanding gass against a mass

• Bipolarity
In summary, the conversation discusses a cylindrical piston containing an ideal gas, with a block on top of it and atmospheric pressure assumed to be negligible. The gas is initially at temperature T and the piston is well insulated, making all thermodynamic processes adiabatic. The cross-section of the piston is A1 and the block has a mass of M and a cross-section of A2. The downward force of Mg on the piston can be interpreted as a pressure of Mg/A2. As the piston expands, the gas's volume increases proportionally to the height of the piston. The gas's pressure decreases and eventually reaches equilibrium with the pressure exerted by the block. The work done by the gas in the expansion is positive, while the work done

#### Bipolarity

Now that I am making progress, I am ready to ask more complex questions
Please correct me where my reasoning is wrong.

Consider a cylindrical piston in which an ideal gas is sealed. The gas is initially at temperature T. The piston is well insulated, so that all thermodynamic processes are at least adiabatic. Assume that the piston is massless and frictionless. Atmospheric pressure is assumed to be negligible.

Now the cross-section of the piston is A1. A block of mass M is placed on the piston and the cross-section of the block is A2. Immediately, due to the weight of the block, a downward force of Mg acts on the piston. This can equally be interpreted as a pressure of Mg/A2 (or is it Mg/A1 ? ).

On the other hand, the gas exerts an outward pressure on the piston much greater than that of the block, so that the piston begins to expand. As the piston expands, the gas's volume increases proportional to the height of the piston, since the cross-section of the piston is a constant.

Now as the gas's volume increases, its pressure must decrease. Eventually, its pressure must reduce to a value equal to the pressure exerted by the block. The piston is now in static equilibrium, since the gas pressure on it equally opposes the pressure due to the block's weight.

My question is, over this process, as the gas expands, can the work it does on the block be calculated? What about the work that the block does on the gas? Which of these is positive, and why?

Is the overall expansion process isothermal? One argument tells me that since the process is adiabatic (by assumption), nonzero work done must be accompanied by nonzero internal energy change (due to the first law), and thus nonzero temperature change (due to the relation between U and T). Thus the temperature must change and the process cannot be isothermal.

Another argument tells me that it is mathematically impossible to calculate the work done by the gas when both temperature and volume change?

So I appreciate all help in this helping understand this concept.

BiP

Bipolarity said:
Now that I am making progress, I am ready to ask more complex questions
Please correct me where my reasoning is wrong.

Consider a cylindrical piston in which an ideal gas is sealed. The gas is initially at temperature T. The piston is well insulated, so that all thermodynamic processes are at least adiabatic. Assume that the piston is massless and frictionless. Atmospheric pressure is assumed to be negligible.

Now the cross-section of the piston is A1. A block of mass M is placed on the piston and the cross-section of the block is A2. Immediately, due to the weight of the block, a downward force of Mg acts on the piston. This can equally be interpreted as a pressure of Mg/A2 (or is it Mg/A1 ? ).
The pressure on the gas is the force per unit area of the block on the gas. The force from the block on the gas is distributed over what area?

On the other hand, the gas exerts an outward pressure on the piston much greater than that of the block, so that the piston begins to expand. As the piston expands, the gas's volume increases proportional to the height of the piston, since the cross-section of the piston is a constant.

Now as the gas's volume increases, its pressure must decrease. Eventually, its pressure must reduce to a value equal to the pressure exerted by the block. The piston is now in static equilibrium, since the gas pressure on it equally opposes the pressure due to the block's weight.

My question is, over this process, as the gas expands, can the work it does on the block be calculated? What about the work that the block does on the gas? Which of these is positive, and why?
One is positive and the other is negative. Work done by the gas in an expansion is positive. Work done on the gas by the surroundings (in this case by the block) in an expansion is negative. The reason has to do with the definition of work as the dot product of a force and the displacement through which the force acts. W = Fd if the directions of force and displacement are the same. W=-Fd if they are in opposite directions. The force of the gas on the mass/piston is in the same direction as the displacement, so the work done by the gas is positive. The force of the Mass/piston on the gas is opposite to the direction of the displacement through which it acts, so the work done by the mass/piston is negative.

Is the overall expansion process isothermal? One argument tells me that since the process is adiabatic (by assumption), nonzero work done must be accompanied by nonzero internal energy change (due to the first law), and thus nonzero temperature change (due to the relation between U and T). Thus the temperature must change and the process cannot be isothermal.
And you would be correct.

Another argument tells me that it is mathematically impossible to calculate the work done by the gas when both temperature and volume change?
It is more difficult. But why is it impossible? It is a just calculus problem.

AM

So the work done by the gas is positive, I get that. It equals the integral of F dot d over the entire interval of displacement.

So what would be the value of F? Would it be equal to the gas's internal pressure times the cross-section of the cylinder, or would it be equal to the weight of the block, times the cross section of the block?

Also, would the work that the block does on the gas just be the negative of the above?

Also, my book uses $$W = -\int^{V2}_{V1}\frac{nRT}{V}dV$$ to calculate the work done. It says that the temperature is constant, and takes it out of the integral.

But I thought we established that the process was not isothermal, so why are we allowed to do this?

BiP

Bipolarity said:
So the work done by the gas is positive, I get that. It equals the integral of F dot d over the entire interval of displacement.

So what would be the value of F? Would it be equal to the gas's internal pressure times the cross-section of the cylinder, or would it be equal to the weight of the block, times the cross section of the block?
It depends on how much greater the gas pressure was than the pressure due to gravity. If the initial gas pressure P is just a little greater than Mg/Area so the piston moves slowly from the beginning, then the pressure in the gas is equal to (Mg + Ma)/Area where a is the acceleration of M. If F is much greater than Mg so that the piston accelerates very rapidly then the gas is not in equilibrium - pressure and temperature are not uniform within the gas so that are undefined for the whole gas.

Also, would the work that the block does on the gas just be the negative of the above?
I think you can work that out. What is the force that the block exerts on the gas? What direction is the displacement?

Also, my book uses $$W = -\int^{V2}_{V1}\frac{nRT}{V}dV$$ to calculate the work done. It says that the temperature is constant, and takes it out of the integral.

But I thought we established that the process was not isothermal, so why are we allowed to do this?
T is not constant in the example you gave of a quasi-static adiabatic expansion. This integral applies to isothermal processes only.

AM

Then what would the integral be for Work in this case when T is not constant ?

I suppose the textbook never really assumed that the process is adiabatic. That is the only way to resolve this issue without violating the first law of thermo and at the same time accept the textbook's mathematics.

BiP

## 1. What is work done by an expanding gas against a mass?

The work done by an expanding gas against a mass is the energy transfer that occurs when a gas expands and exerts a force on a mass, causing it to move.

## 2. How is work done by an expanding gas against a mass calculated?

Work done by an expanding gas against a mass is calculated by multiplying the force exerted by the gas on the mass by the distance the mass moves in the direction of the force.

## 3. What factors affect the work done by an expanding gas against a mass?

The work done by an expanding gas against a mass is affected by the initial and final volumes of the gas, the pressure of the gas, and the mass and distance of the object being moved.

## 4. How does work done by an expanding gas affect the temperature of the gas?

Work done by an expanding gas against a mass can cause a decrease in the temperature of the gas due to the decrease in internal energy of the gas as it expands and does work against the mass.

## 5. Can work done by an expanding gas be converted into other forms of energy?

Yes, work done by an expanding gas can be converted into other forms of energy, such as heat or mechanical energy, depending on the situation and the type of work being done.