Question

A reversible cyclic process for an ideal gas is shown below. Here, P, V and T are pressure, volume and temperature, respectively. The thermodrynamic parameteres q, w and U are heat, work and internal energy respectively.
The incorect option(s) is(are):

• A. $w_{AB}=-P_2(V_2-V_1)$
  
• B. $w_{BC}=-P_2(V_2-V_1)$
• C. $\Delta U_{AB}=0$, $\Delta U_{CA}=q_{CA}$
• D. $w_{CA}=0$

Answer 1

Answer: (A), (B)

$w_{BC}=-P_2(V_2-V_1)$

Process AB is both isothermal and reversible so,
$w_{AB}=nR{T}_{1}ln\left(\frac{V_2}{V_1}\right)$
hence, A is incorrect

Process BC is isobaric so,
$w_{BC}=-P_2(V_1-V_2)$
hence B is incorrect

Process AB is isothermal so,
$\Delta U_{AB}=0$

Process CA is isochoric so, $\Delta V=0$
hence, $w_{CA}=0$
So, using the first law, we get
$\Delta U_{CA}=q_{CA}$
so, option C,D are correct

Hence options a and b are incorrect.