Question

A gas is taken through a cyclic process ABCA as shown in figure. If 2.4 cal of heat is given in the process, what is the value of J ?

Answer 1

Heat given in the process, ∆Q = 2.4 cal

∆W = W_AB + W_BC +  W_CA

For line AB:-

Change in volume,

ΔV = 0

∴ W_AB = 0

For line BC :-

Mean pressure, \[P = \frac{1}{2} \times (100 + 200) kPa = 150 \times {10}^3 Pa\]

\[ ∆ V = (700 - 500) cc = 200 cc\]

\[ W_{BC} = 150 \times {10}^3 \times 200 \times {10}^{- 6} \]

\[ W_{BC} = 30 J\]

For line AC :-

P = 100 kPa

ΔV = 200 cc

W_CA = 100 × 10³ × 200 × 10⁻⁶

Total work done in the one cycle is given by

∆W = 0 + 150× 10³ × 200 × 10⁻⁶ − 100 × 10³ × 200 × 10⁻⁶

∆W = `1/2` × 300 × 10³ × 200 × 10⁻⁶ − 20

∆W = 30 J − 20 J = 10 J

∆U = 0    .............(in a cyclic process)

∆Q = ∆U + ∆W

⇒ 2.4J = 10

\[\Rightarrow J = \frac{10}{2 . 4} = \frac{100}{24} = \frac{25}{6} = 4 . 17 \text{J/cal}\]