Question

A diatomic ideal gas undergoes a thermodynamic change according to the $P-V$ diagram shown in the figure. The total heat given to the gas is nearly (use $ln2=0.7$ )

• A. $2.5P_0{V}_0$
• B. $1.4P_0{V}_0$
• C. $39P_0{V}_0$
• D. $1.1P_0{V}_0$

Answer 1

Answer: (C)

$Q_{AB}=\Delta U_{AB}+W_{AB}$ , where $W_{AB}=0$
$\Delta U_{AB}=\frac{f}{2}nR\Delta T$
$\Rightarrow \Delta U_{AB}=\frac{f}{2}\left(\Delta PV\right)$
$\Rightarrow \Delta U_{AB}=\frac{5}{2}\left(\Delta PV\right)$
$\Rightarrow \Delta U_{AB}=2.5\left(\Delta PV\right)$
According to the first law of thermodynamics $Q_{AB}=\Delta U_{AB}+W_{AB}$
Where $W_{AB}=0$
therefore $Q_{AB}=\Delta U_{AB}$
$\Rightarrow Q_{AB}=2.5P_0{V}_0$

Process $BC$
$Q_{BC}=\Delta U_{BC}+W_{BC}$
$\Rightarrow Q_{BC}=0+2P_0{V}_{0}ln2$
$Q_{BC}=1.4P_0{V}_0$
$Q_{net}=Q_{AB}+Q_{BC}=3.9P_0{V}_0$