How Does Non-Quasistatic Compression Affect Gas in a Sealed Cylinder?

[superspartan9]

Homework Statement

A cylinder contains one liter of air and room temperature (300 K) and atmospheric pressure (10⁵ N/m²). At one end of the cylinder is a massless piston, whose surface area is 0.01m². Suppose that you push the piston in VERY suddenly, exerting 2000N. The piston moves only one millimeter before it is stopped by an immovable barrier of some sort.

a) How much work have you done on this system?
b) How much heat has been added to the gas?
c) Assuming that all the energy added goes into the gas (not the piston or cylinder walls), by how much does the internal energy of the gas increase?
d) Use the thermodynamic identity to calculate the change in the entropy of the gas (once it has again reached equilibrium.

Homework Equations

dW = -P dV
Q = T dS (quasistatic)
dS > Q/T when W > -P dV --> in text for example problem much like this.
dU = T dS - P dV (thermodynamic identity)
ΔU = Q + W

The Attempt at a Solution

My big issue right now is that the equations presented in the chapter for this scenario are > / < not =, therefore, how can I know how much work and heat go into the system??
Without that, I just go through the numbers as usual with W = P * dV = 2000 N / 0.01 m² * (0.001 m) * (0.01 m²), which gives me a solid number.
Then I believe this process is adiabatic, therefore, the Q = 0
Which then makes c) really easy since it's just work going into the system.
d) then becomes a simply math problem.
Is this right? Is the problem supposed to be this simple?

 

[mark726]

Hello, thank you for your question. Let's break down each part of the problem to ensure we have a complete understanding of the scenario.

a) How much work have you done on this system?

To calculate the work done on the system, we can use the equation dW = -P dV, where dV is the change in volume and P is the pressure. In this case, the change in volume is 0.001 m3 (1 millimeter = 0.001 meters) and the pressure is 105 N/m2. Plugging these values into the equation, we get dW = -(105 N/m2)(0.001 m3) = -0.105 J. Therefore, you have done -0.105 J of work on the system.

b) How much heat has been added to the gas?

To calculate the heat added to the gas, we can use the equation Q = T dS (quasistatic), where T is the temperature and dS is the change in entropy. Since the temperature remains constant at 300 K, we can simplify the equation to Q = TΔS. Since we know that the process is adiabatic (no heat exchange), Q = 0. Therefore, no heat has been added to the gas.

c) Assuming that all the energy added goes into the gas (not the piston or cylinder walls), by how much does the internal energy of the gas increase?

Since the process is adiabatic and there is no heat exchange, the change in internal energy (ΔU) is equal to the work done on the system (dW). Therefore, the internal energy of the gas increases by -0.105 J.

d) Use the thermodynamic identity to calculate the change in the entropy of the gas (once it has again reached equilibrium).

The thermodynamic identity is given by dU = T dS - P dV. Rearranging this equation, we get dS = (dU + P dV)/T. We know that the change in internal energy (dU) is -0.105 J and the change in volume (dV) is 0.001 m3. Plugging these values into the equation, we get dS = (-0.105 J + (105 N/m2)(0.001 m3))/300 K = 3.5 x 10^-7 J/K.