Question

An ideal gas goes through a reversible cycle $a\to b\to c\to d$ has the V- T diagram shown below. Process $d\to a$ and $b\to c$ are adiabatic.
The corresponding P-V diagram for the process is ( all figure are schematic and not drawn to scale).

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

In process $a\to b$ and $c\to d$ the volume is directly proportional to temperature of gas, hence the process is isobaric, however, the slope of curve is different in each case. using ideal gas equation

$PV=nRT\Rightarrow V=\frac{nR}{P}T$

slope $m=\frac{nR}{P}$ for a high pressure the slope of the V-T curve will be less. In given diagram the slope of $c\to d$ process is lower than $a\to b$ process, hence process $a\to b$ is at higher pressure, from which option B and D can be ruled out.

Now initial rate of change of volume w.r.t. temperature is very less from $b\to c$ and $d\to a$ as shown in V-T diagram.

hence slope of P-V curve should tend to ${lim}_{\Delta V\to 0}\frac{\partial P}{\partial v}=\infty$ as initial change in volume is nearly zero, which can be verified in option A

Hence correct answer is option A.