A cylinder contains one liter of air at room temperature (300 K) and atmospheric pressure (10^5 N/m^2). At one end of the cylinder is a massless piston, whose surface area is 0.01 m^2. Suppose that you push the piston in very suddenly, exerting a force of 2000 N. 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).
The thermodynamic identity is:
dU= TdS-PdV
For fast compression dS > Q/T (when W>-PdV)
a) Well, pushing a piston sounds like a lot of work! So, the answer must be "A ton of work!" But in all seriousness, to calculate the work done on the system, we can use the formula: work = force × distance. Since the force exerted is 2000 N and the distance moved is 0.001 m (1 millimeter), the work done is 2000 N × 0.001 m = 2 J (Joules).
b) As for the heat added to the gas, we need a little equation called the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added minus the work done on the system. Since the work done is 2 J and there is no mention of heat added, we can say that the heat added is equal to the work done, which is 2 J.
c) Ah, the internal energy of the gas! Well, when work is done on a system, the internal energy can increase. In this case, the internal energy of the gas increases by the amount of work done on the system, which is 2 J.
d) Now, let's talk about entropy, shall we? Entropy is like a measure of disorder. When a gas is compressed quickly, there is an increase in entropy. But to calculate the change in entropy, we need the final and initial states of the gas. Since we don't have those values, I'm afraid we can't calculate the change in entropy. But hey, at least we had some fun along the way, right?
To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. We can break down the problem into separate parts to determine the answers to each question.
a) To find the work done on the system, we can use the formula:
Work = Force x Distance.
The force exerted on the piston is 2000 N, and the distance the piston moves is 0.001 m (since 1 millimeter is equal to 0.001 meters). Plugging these values into the formula, we get:
Work = 2000 N x 0.001 m = 2 J.
Therefore, you have done 2 Joules of work on the system.
b) To find the heat added to the gas, we can use the formula:
Heat = Change in Internal Energy + Work.
Since the piston is stopped by an immovable barrier, the volume of the gas remains constant. Therefore, from the first law of thermodynamics, we know that the change in internal energy of the gas is equal to the work done on the system. Thus, the heat added to the gas is:
Heat = Work = 2 J.
So, 2 Joules of heat have been added to the gas.
c) The internal energy of the gas increases by the same amount as the heat added to the gas. Therefore, the internal energy of the gas increases by 2 J.
d) To calculate the change in entropy of the gas, we can use the thermodynamic identity: dU = TdS - PdV.
Since the volume of the gas remains constant, dV = 0.
Therefore, the equation simplifies to: dU = TdS.
Rearranging the equation, we get: dS = dU / T.
Using the values from previous calculations, dU = 2 J and T = 300 K (room temperature).
Plugging these values into the equation, we get:
dS = 2 J / 300 K = 0.0067 J/K.
Thus, the change in entropy of the gas is 0.0067 J/K (once it has again reached equilibrium).
Note: The given statement dS > Q/T for fast compression (W > -PdV) is not directly applicable in this problem since we have a fixed volume. However, it is true in general for fast compression processes.