Question

The shown p- V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle is

• A. $p_0v_0$
• B. $\bigg(\frac{13}{2}\bigg)p_0v_0$
• C. $\bigg(\frac{11}{2}\bigg)p_0v_0$
• D. $4p_0v_0$

Answer 1

Answer: (B)

$\bigg(\frac{13}{2}\bigg)p_0v_0$

Answer 2

Heat is extracted from the source means heat is given to the system (or gas) or Q is positive. This is positive only along the path ABC.
Heat supplied
$\therefore \, \, \, \, Q_{ABC}+W_{ABC}$
$\, \, \, \, \, \, \, \, \, \, \, =nC_v(T_f -T_i)+$ Area under p-V graph
$\, \, \, \, \, \, \, \, \, \, \, =n\bigg(\frac{3}{2}R\bigg)(T_C-T_{A})+2p_0v_0$
$\, \, \, \, \, \, \, \, \, \, =\frac{3}{2}(nRT_C-nRT_A)+2p_0v_0$
$\, \, \, \, \, \, \, \, \, \, =\frac{3}{2} (p_C V_C-p_A V_A)+2p_0V_0$
$\, \, \, \, \, \, \, \, \, \, =\frac{3}{2} (4p_0V_0-p_0V_0)+2p_0V_0$
$\, \, \, \, \, \, \, \, \, \, =\bigg(\frac{13}{2}\bigg)p_0V_0$