A monoatomic ideal gas initially at temperatu | vedclass

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(C) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Therefore,$\frac{T_

Vedclass · Accepted Answer

$\left(\frac{L_1}{L_2}\right)^{2/3}$

Vedclass · Answer

(C) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Therefore,$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}$.For a monoatomic ideal gas,the adiabatic index $\gamma = \frac{5}{3}$.Thus,$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.Since the gas is in a cylinder of constant cross-sectional area $A$,the volume $V$ is proportional to the length $L$ of the gas column $(V = AL)$.Substituting $V_1 = AL_1$ and $V_2 = AL_2$,we get $\frac{V_1}{V_2} = \frac{L_1}{L_2}$.Substituting these values into the temperature ratio equation:$\frac{T_2}{T_1} = \left(\frac{L_1}{L_2}\right)^{2/3}$.

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