# Work done by gas expanding into a cylinder+piston system

• LCSphysicist
In summary, the reasoning for both letters is briefly summarized as follows: - Letter A: The work is quasi-static, with a constant pressure equal to atmospheric pressure. Therefore, the work is calculated as W = -p∫dV = -p0(V0-V). - Letter B: The work is fast, with no heat flow through the gas. The work is calculated as W = -∫pdV = -∫ViVfpi(Vi)γdV/Vγ. PiViγ / Po = Vfγ. The gas is still leaking slowly into the atmosphere instead of into a cylinder and piston. The final points in (PV space) may be the same, but the path could be
LCSphysicist
Homework Statement
A thin-walled metal container of volume V contains a gas at high pressure.
Connected to the container is a capillary tube and stopcock. When the stopcock is
opened slightly, the gas leaks slowly into a cylinder equipped with a nonleaking,
frictionless piston, where the pressure remains constant at the atmospheric value $P_0$.
(a) Show that, after as much gas as possible has leaked out, an amount of work $P_0(V_0-V)$
has been done, where $V_0$ is the volume of the gas at atmospheric pressure and
temperature.
(b) How much work would be done if the gas leaked directly into the atmosphere?
Relevant Equations
.
I will summarize briefly my reasoning for both letters, since the answer is immediately after that:

A) The work is quasi-static, and the pressure is approximatelly constant and equal to the atmospheric pressure, so the works is $$W = -p\int dV = -p_{0} (V_{0}-V)$$
B) The work is fast, fast enough that no heat flows thought the gas, so that the work is $$W = -\int p dV = -\int_{V_i}^{V_f} p_i (V_i)^{\gamma} dV /V^{\gamma}$$ $$P_i V_i^{\gamma} / P_o = V_f ^{\gamma}$$

What do you think? I am almost sure it is wrong, so i am posting here to get any help on letter b

If your system is the gas, why do you feel that the answer to B is different from A?

Chestermiller said:
If your system is the gas, why do you feel that the answer to B is different from A?
As i said, because the first one "leaks slowly", and nothing has been said about the second one, so i suppose it is more "explosive".

I have questioned myself also if the work is the same, but i was not able to argue why it should be. Even so the final points on (PV space) is the same, the path could be different, resulting in different work.

The question implies that the gas is still leaking slowly in case B, but into the atmosphere instead of into a cylinder and piston.

## 1. What is work done by gas expanding into a cylinder+piston system?

The work done by gas expanding into a cylinder+piston system is the force exerted by the gas on the piston multiplied by the distance the piston moves. This is known as mechanical work and is measured in joules (J).

## 2. How is work done by gas expanding into a cylinder+piston system calculated?

The work done can be calculated using the formula W = F * d, where W is the work done, F is the force exerted by the gas on the piston, and d is the distance the piston moves.

## 3. Can the work done by gas expanding into a cylinder+piston system be negative?

Yes, the work done can be negative if the gas is compressed and the piston moves in the opposite direction. This means that the gas is doing work on the surroundings instead of the surroundings doing work on the gas.

## 4. What factors affect the work done by gas expanding into a cylinder+piston system?

The work done is affected by the initial and final volumes of the gas, the pressure of the gas, and the temperature of the gas. These factors determine the amount of force the gas exerts on the piston and the distance the piston moves.

## 5. What is the relationship between work done and energy in a gas expanding into a cylinder+piston system?

The work done by the gas expanding into a cylinder+piston system is equal to the change in energy of the gas. This means that the work done is directly proportional to the change in energy, and can be used to calculate the change in internal energy of the gas.

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