Question

Consider the following cyclic process carried out in two steps on a gas.

Step 1: 35 J of heat is added to the gas, and 12 J of expansion work is performed.
Step 2: 65 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. J

Answer 1

`W_2=18J`

Step-by-step explanation:

Hello,

In this case, since this is a cyclic process, the involved energy change is the same for the first process and the second process, Thus, with the added heat and work done by the system, we compute such energy change for the first process:

`\Delta E=Q_1-W_1=35J+12J=47J`

Now, for the second process, in order to take the system to the initial state, the same energy change is present, therefore, the needed work turns out:

`W_2=Q_2-\Delta E=65J-47J\\W_2=18J`

Best regards.

Answer 2

The work for the gas compression is 42 J

Step-by-step explanation:

Step 1: Data given

35 J of heat is added to the gas

12 J of expansion work

65 J of heat is removed from the gas as the gas is compressed back to the initial state

Step 2:

ΔE = q + w

⇒with q = 35 J of heat

⇒with w = 12 J of expansion work

⇒ expansion work = work done by the gas = w is negative

ΔE = 35 J - 12 J

ΔE = 23 J

Step 3: Calculate the work for the gas compression

ΔE = -23 J = -65 J + w

⇒65 J of heat is removed from the gas as the gas is compressed back to the initial state

⇒ We need 23 J to go back to the initial state: ΔE = -23 J

Compressed gas means work done by the surroundings => w is positive

w = -23 J + 65 J = 42 J

The work for the gas compression is 42 J