A monoatomic ideal gas, initially at temperat | vedclass

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(C) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.For a monoatomic i

Vedclass · Accepted Answer

$\left(\frac{l_{2}}{l_{1}}\right)^{\frac{2}{3}}$

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(C) For an adiabatic process, the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.For a monoatomic ideal gas, the adiabatic index $\gamma = \frac{5}{3}$.Thus, the relation becomes $T_{1}V_{1}^{\gamma-1} = T_{2}V_{2}^{\gamma-1}$.Rearranging for the temperature ratio: $\frac{T_{1}}{T_{2}} = \left(\frac{V_{2}}{V_{1}}ight)^{\gamma-1}$.Since the gas is in a cylinder of cross-sectional area $A$, the volume $V = A 	imes l$, where $l$ is the length of the gas column.Substituting $V_{1} = A l_{1}$ and $V_{2} = A l_{2}$, we get $\frac{V_{2}}{V_{1}} = \frac{l_{2}}{l_{1}}$.Substituting $\gamma = \frac{5}{3}$, we get $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.Therefore, $\frac{T_{1}}{T_{2}} = \left(\frac{l_{2}}{l_{1}}ight)^{\frac{2}{3}}$.

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