Question

An ideal gas is taken through the cycle A $\to$ B $\to$ C $\to$ A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C $\to$ A is

• A. -5 J
• B. -10 J
• C. -15 J
• D. -20 J

Answer 1

Answer: (A)

$\Delta W_{AB}=p\Delta V=(10)(2-1)=10J$
$\Delta W_{BC}=0$ (as V = constant)
From first law of thermodynamics
$\Delta Q=\Delta W+\Delta U$
$\Delta U=0$ (process ABCA is cyclic)
$\therefore \Delta Q=\Delta W_{AB}+\Delta W_{BC}+\Delta W_{CA}$
$\therefore \Delta W_{CA}=\Delta Q-\Delta W_{AB}-\Delta W_{BC}$
$=5-10-0=-5J$