Chemistry - Thermodynamics - I Question with Solution | TestHub

First Law Relation: From $q=\Delta U+W$ and the given condition $q=-\Delta U$, the work done by the gas is $W=-2\Delta U=-2n{C}_v(T_f-T_i)$.

Polytropic Index ($x$): Using the molar heat capacity formula $C=C_v+\frac{R}{1-x}$ with $C=-C_v=-1.5R$ yields:

$$-1.5R=1.5R+\frac{R}{1-x}$$

This simplifies to $x=\frac{4}{3}$.

Final Temperature ($T_f$): Since $T{V}^{x-1}=constant$, we have $T_i{V}_i^{1/3}=T_f(8V_i{)}^{1/3}$, which simplifies to $T_f=\frac{T_i}{2}=\frac{300}{2}=150K$.

Work Done Value: Substituting the values into the work equation gives $W=-2\times 1\times (1.5\times 2)\times (150-300)=900Cal$ (Option A).