Consider the following cyclic process carried out in two steps on a gas.
Step 1: 41. J of heat is added to the gas, and 20. J of expansion work is performed.
Step 2: 60. J of heat is removed from the gas as the gas is compressed back to the initial state.
Calculate the work for the gas compression in Step 2.
To calculate the work for the gas compression in Step 2, 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 are given the following information:
- In Step 1, 41 J of heat is added to the gas, and 20 J of expansion work is performed.
- In Step 2, 60 J of heat is removed from the gas as the gas is compressed back to the initial state.
Now, let's calculate the work for the gas compression in Step 2.
Since we have the heat transfer and we want to find the work done, we can use the first law of thermodynamics to express the change in internal energy:
ΔU = Q - W
Where:
ΔU is the change in internal energy
Q is the heat added to the system
W is the work done by the system
In Step 1, the net heat added to the system can be calculated as follows:
Q1 = Q_in - Q_out
= 41 J - 0 J
= 41 J
The work done by the system in Step 1 is given as 20 J.
Therefore, the change in internal energy in Step 1 can be calculated as:
ΔU1 = Q1 - W1
= 41 J - 20 J
= 21 J
Now, in Step 2, the heat removed from the system is given as 60 J. Therefore, we can write:
Q2 = -60 J
We want to find the work done by the system in Step 2, so we can rearrange the first law of thermodynamics equation as follows:
ΔU2 = Q2 - W2
⇒ W2 = Q2 - ΔU2
Substituting the values, we get:
W2 = -60 J - 21 J
= -81 J
Therefore, the work done for the gas compression in Step 2 is -81 J (negative sign indicates work done on the gas).
Note: The negative sign indicates that work is done on the gas during compression, rather than being done by the gas.