Answer:
the work for the gas compression in step 2 = 32 J
Explanation:
Given that:
A system usually comes back to initial state after 2 steps;
That implies :
ΔE₁ + ΔE₂ = 0
When heat is added to a given system , q tends to be positive
q is also negative when the heat is removed from the system
The work (W) when expansion occurs is said to be negative and positive when compression occurs.
∴ ΔE₁ = q₁ + W₁
ΔE₁ = 50 J + (-20 J)
ΔE₁ = 30 J
ΔE₂ = q₂ + W₂
ΔE₂ = -62 J + W₂
ΔE₁ + ΔE₂ = 0
30 J -62 J + W₂ = 0
W₂ = - 30 J + 62 J
W₂ = 32 J
Thus, the work for the gas compression in step 2 = 32 J
Answer:
The work for the gas compression is 32 J
Explanation:
Step 1: Data given
50 J of heat is added to the gas
20 J of expansion work
62 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 = 50 J of heat
⇒with w = 20 J of expansion work
⇒ expansion work = since there is work done by the gas: w is negative
ΔE = q + w
ΔE = 50 J - 20 J
ΔE = 30J
Step 3: Calculate the work for the gas compression
ΔE = -30 J = -62 J + w
⇒62 J of heat is removed from the gas as the gas is compressed back to the initial state. Compressed gas means work done by the surroundings => w is positive
⇒To go back to the initial state, we need 30 J
ΔE = -23 J
w = -30 J + 62 J = 32 J
The work for the gas compression is 32 J
