Question

An ideal gas is taken through the cycle A $\rightarrow$ B $\rightarrow$ 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 $\rightarrow$ A is

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

Answer 1

Answer: (A)

$\Delta W_{AB}=p\Delta V =(10)(2-1)=10 J$
$ \, \, \, \, \, \, \, \, \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=-5 J$