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.
Figure

Answer 1

50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB. The quantity of heat to be supplied to take it from A to B via ADB  = 55 cal

Explanation:

In ACB  path,

∆Q = 50 cal = (50 × 4.2) J

∆Q = 210 J

$\Delta W = W_{AC} + W_{CB}$

Because initial and final volumes run along line BC are the same, Variations in unit volume in BC are small.

Therefore work done on this line is going to be zero.

For AC  line:

P = 50 k Pa  

Changes  Volume (200 cc to  400 cc).

ΔV = 400 − 200 cc = 200  

$\Delta W = W_{AC} + W_{CB}$

     = 50 × 10 − 3 × 200 × 10 − 6 + 0

     = 10 J

Using the thermodynamics first law, we get

∆Q = ∆U + ∆W

⇒ ∆U = ∆Q − ∆W = (210 − 10) J

∆U = 200 J

In ADB  path, ∆Q = ?

∆U = 200 J            

$\Delta W = W_{AD} + W_{DB} =\Delta W$

Work done for AD  line will also be 0.  

For DB line:

P = 155 k Pa  

ΔV = 400 - 200 = 200  

$W = 0 + 155 \times10^3 \times 200 \times10^{-6}$

W = 31 J

∆Q = ∆U + ∆W

∆Q = (200 + 31) J = 231 J

∆Q = 55 cal