Adiabatic Process

Question

A monatomic ideal gas with gamma equal to 5/3 expands adiabatically until its volume becomes 8 times the initial value, so what is the ratio of final to initial temperature?

RankGuru · Accepted Answer

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 $PV^{\gamma} = 	ext{constant}$, which combined with the ideal gas law gives the temperature-volume form $TV^{\gamma-1} = 	ext{constant}$. Applying this between the two states, $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$, and rearranging yields $\dfrac{T_2}{T_1} = \left(\dfrac{V_1}{V_2}ight)^{\gamma-1}$. 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 $\dfrac{V_1}{V_2} = \dfrac{1}{8}$. Therefore $\dfrac{T_2}{T_1} = \left(\dfrac{1}{8}ight)^{2/3} = (8^{-1})^{2/3} = 8^{-2/3} = (2^3)^{-2/3} = 2^{-2} = \dfrac{1}{4}$. 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.

Adiabatic Process

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