Question

Two moles of triatomic linear gas are taken through a reversible process starting from A as shown in figure. The volume ratio $\frac{V_B}{V_A}=4$ and $\frac{V_C}{V_A}=16$. If the temperature at 'A' is $-{73}^{\circ }C$ then select
correct option: [Use : ln 2 = 0.7]

• A. $W_{AC}=-1440R$
• B. $\Delta H_{AB}=4200R$
• C. $\Delta H_{BC}=4200R$
• D. $q_{AC}=4200R$

Answer 1

The correct option is B $\Delta H_{AB}=4200R$

Moles of triatomic gas=2
$\frac{V_B}{V_A}=4$ and $\frac{V_C}{V_A}=16$
$T_A=-{73}^{\circ }C$
$\frac{T_B}{V_B}=\frac{T_A}{V_A}$

$T_B=\frac{T_A\times V_B}{V_A}=4\times 200=800K$
$\Delta H_{AC}=\Delta H_{AB}+\Delta H_{BC}\Delta H_{AC}=nCp\Delta T+0=2\times \frac{7}{2}\times R\times (1800-200)=4200R$
$\Delta U_{AC}=\Delta U_{AB}+\Delta U_{BC}\Delta U_{AC}=nCv\Delta T+0=2\times \frac{5}{2}\times R\times (800-200)=3000R$
$W_{AB}=-nR\Delta T$
$W_{AB}=-2R600=-1200R$
$W_{BC}=-2R\times 800ln4=-2218.07R$
$\Rightarrow W_{AC}=-3418.07R$
From first law of thermodynamic
$\Rightarrow q_{AC}=\Delta U_{AC}-W_{AC}=3000R+3440R=6440R$
Hence option (b) is correct.