A monoatomic ideal gas,initially at temperatu | vedclass

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(A) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For a monoatomic ideal gas,the d

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

Vedclass · Answer

(A) For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For a monoatomic ideal gas,the degrees of freedom $f = 3$.The adiabatic index $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{3} = \frac{5}{3}$.Thus,$\gamma - 1 = \frac{2}{3}$.Since the volume $V$ is proportional to the length $L$ of the gas column ($V = A 	imes L$,where $A$ is the constant cross-sectional area),we have $\frac{V_1}{V_2} = \frac{L_1}{L_2}$.Substituting this into the adiabatic relation: $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}ight)^{\gamma-1} = \left(\frac{L_1}{L_2}ight)^{2/3}$.

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