One mole of an ideal gas in initial state A undergoes a cyclic process ABCA, as shown in Fig. Its pressure at A is $P_0$. Choose the correct option(s) from the following.
(A) Internal energies at A and B are the same
(B) Work done by the gas in process AB is $P_0 V_0 ln 4 $
(C) Pressure at C is $P_0/4 $
(D) Temperature at C is $T_0/4$
Question source: JEE advanced 2010
I have doubt only for option (C) and (D). You may attempt question with neutral mind before opening the spoilers.
Official answer: (A),(B),(C),(D)
My answer: (A),(B)
Method: There is no information of temperature or pressure at point C. From diagram information is revealed that process BC is linear but it may or may not be isobaric. So, we cannot conclude or comment anything about option C and D. Hence, they should not be ticked in exam. Is there any condition that process BC must be isobaric?
In general: If the gas follows a line in V-T space, must the process be isobaric?
Best Answer
There is a difference between proportionality and a more general straight line. I get the equation for the line as $$V=V_0\left[1+3\frac{(T-T_C)}{(T_0-T_C)}\right]$$So, $$P=\frac{P_0(T/T_0)}{\left[1+3\frac{(T-T_C)}{(T_0-T_C)}\right]}$$The only way that P is the same at the two end points B and C is if $T_C=T_0/4$. Mathematically, this will also guarantee that P is constant at all other points along the line.