A monoatomic ideal gas,initially at temperatu | vedclass

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Question

(A) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.For a monoatomic id

Vedclass · Accepted Answer

$\left[\frac{L_1}{L_2}\right]^{2/3}$

Vedclass · Answer

(A) For an adiabatic process,the relationship between temperature and volume is given by $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.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 $V$ of the gas column is proportional to its length $L$ (as $V = A 	imes L$ and area $A$ is constant),we have $\frac{V_2}{V_1} = \frac{L_2}{L_1}$.Substituting this into the adiabatic equation: $\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}ight)^{\gamma-1} = \left(\frac{L_1}{L_2}ight)^{2/3}$.

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