Question

The above 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. $P_o{V}_o$
• B. $\left(\frac{13}{2}\right)P_o{V}_o$
• C. $\left(\frac{11}{2}\right)P_o{V}_o$
• D. $4P_o{V}_o$

Answer 1

The correct option is B $\left(\frac{13}{2}\right)P_o{V}_o$

Total heat taken $\left(Q_h\right)$ = $(n{C}_v\Delta T{)}_{A\to B}+(n{C}_p\Delta T{)}_{B\to C}$
Since monoatomic gas is given hence $C_v=\frac{3R}{2}$ and $C_p=\frac{5R}{2}$
$Q_h={\left(n\frac{3R}{2}\Delta T\right)}_{A\to B}+{\left(n\frac{5R}{2}\Delta T\right)}_{B\to C}$
$Q_h={\left(\frac{3}{2}V\Delta P\right)}_{A\to B}+{\left(\frac{5}{2}P\Delta V\right)}_{B\to C}$
$Q_h=\frac{13}{2}{P}_o{V}_o$