A monoatomic ideal gas is compressed adiabati | vedclass

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

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$9$

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(B) 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 $\gamma = \frac{5}{3}$.Therefore, $\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.Given the initial state $(T_1, V_1) = (T, V)$ and the final state $(T_2, V_2) = (xT, V/27)$.Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.$T(V)^{2/3} = (xT) \left(\frac{V}{27}ight)^{2/3}$.$1 = x \left(\frac{1}{27}ight)^{2/3}$.$1 = x \left(\left(\frac{1}{3^3}ight)^{1/3}ight)^2 = x \left(\frac{1}{3^2}ight) = x \left(\frac{1}{9}ight)$.Thus, $x = 9$.

$A$ monoatomic ideal gas is compressed adiabatically to $\left(\frac{1}{27}\right)$ of its initial volume. If the initial temperature of the gas is $T \ K$ and the final temperature is $xT \ K$, the value of $x$ is:

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