Adiabatic Process

Per NCERT Class 11, Chapter 12 (Thermodynamics), an adiabatic process exchanges no heat with the surroundings, and for an ideal gas the state variables are linked by $P{V}^{\gamma }=\text{constant}$, which combined with the ideal gas law gives the temperature-volume form $T{V}^{\gamma -1}=\text{constant}$. Applying this between the two states, $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$, and rearranging yields $\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$. For a monatomic gas the ratio of specific heats is $\gamma =5/3$, so the exponent is $\gamma -1=2/3$. Here the gas expands to $V_2=8V_1$, hence $\frac{V_1}{V_2}=\frac{1}{8}$. Therefore $\frac{T_2}{T_1}={\left(\frac{1}{8}\right)}^{2/3}=(8^{-1}{)}^{2/3}=8^{-2/3}=(2^3{)}^{-2/3}=2^{-2}=\frac{1}{4}$. The option 1/8 is wrong because it uses an exponent of 1 instead of $\gamma -1=2/3$. The option 1/2 corresponds to $8^{-1/3}$, which uses the wrong power. The option 1/16 corresponds to $8^{-4/3}$, an exponent error that doubles the correct value. A plausibility check confirms the answer: adiabatic expansion does work at the expense of internal energy and so cools the gas, meaning $T_2<T_1$, and indeed $1/4<1$, while the moderate exponent $2/3$ keeps the temperature drop gentler than the eightfold volume increase, exactly as the result shows.