Question

Consider the following cyclic process carried out in two steps on a gas. Step 1: 50. J of heat is added to the gas, and 20. J of expansion work is performed. Step 2: 62 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.

Answer 1

The work for the gas compression is 32 J

Step-by-step 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

Answer 2

the work for the gas compression in step 2 = 32 J

Step-by-step 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