Question

50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in figure. Find the quantity of heat to be suppled to take it from A to B via ADB.

Answer 1

Given:
In path ACB,
∆Q = 50 cal = (50 × 4.2) J
∆Q = 210 J
∆W = W_AC + W_CB
Since initial and final volumes are the same along the line BC, change in volume of the system along BC is zero.
Hence, work done along this line will be zero.

For line AC:
P = 50 kPa

Volume changes from 200 cc to 400 cc.
$\Rightarrow ∆V=400-200cc=200cc$
∆W = W_AC + W_CB
= 50 × 10⁻³ × 200 × 10⁻⁶ + 0
= 10 J

Using the first law of thermodynamics, we get
∆Q = ∆U + ∆W
⇒ ∆U = ∆Q − ∆W = (210 − 10) J
∆U = 200 J

In path ADB, ∆Q = ?
∆U = 200 J (Internal energy depends only on the initial and final points and not on the path followed.)

⇒ ∆W = W_AD + W_DB = ∆W

Work done for line AD will also be zero.

For line DB:
P = 155 kPa
$∆V=400-200=200cc$
W = 0 + 155 × 10³ × 200 × 10⁻⁶
W = 31 J

∆Q = ∆U + ∆W
∆Q= (200 + 31) J = 231 J
∆Q = 55 cal