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)

$Q_{A B}=\Delta U_{A B}+W_{A B}$ , where $W_{A B}=0$
$\Delta U_{A B}=\frac{f}{2}nR\Delta T$
$\Rightarrow \Delta U_{A B}=\frac{f}{2}\left(\Delta P V\right)$
$\Rightarrow \Delta U_{A B}=\frac{5}{2}\left(\Delta P V\right)$
$\Rightarrow \Delta U_{A B}=2.5\left(\Delta P V\right)$
According to the first law of thermodynamics $Q_{A B}=\Delta U_{A B}+W_{A B}$
Where $W_{A B}=0$
therefore $Q_{A B}=\Delta U_{A B}$
$\Rightarrow Q_{A B}=2.5P_{0}V_{0}$

Process $BC$
$Q_{B C}=\Delta U_{B C}+W_{B C}$
$\Rightarrow \, \, \, Q_{B C}=0+2P_{0}V_{0}ln2$
$Q_{B C}=1.4P_{0}V_{0}$
$Q_{n e t}=Q_{A B}+Q_{B C}=3.9P_{0}V_{0}$