Question

An ideal monoatomic gas undergoes a cyclic process ABCA as shown in the figure.Heat absorbed during AB is :

• A. $\frac{5R{T}_O}{2}$
• B. $\frac{5}{3}$
• C. $\frac{5}{4ln2}$
• D. $\frac{5}{6}$

Answer 1

The correct option is C $\frac{5}{4ln2}$

$\text{Solution}:$

$W_{AB}=(2V_0-V_0)P_0=P_0{V}_0$

[isobaric process]

$W_{BC}=|nRT\left(2T_0\right)ln\frac{V_0}{2V_0}|=2P_0{V}_{0}ln2$

[Isothermal process]

$\therefore \frac{Q_{AB}}{W_{BC}}=\frac{\frac{3}{2}{P}_0{V}_0+P_0{V}_0}{2P_0{V}_{0}ln2}=\frac{5}{4ln2}$

$\text{Hence C is the correct option}$