A diatomic ideal gas is compressed adiabatica | 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 diatomic gas,

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$4$

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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 diatomic gas,the adiabatic index $\gamma = \frac{7}{5}$.Thus,$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Given $V_2 = \frac{V_1}{32}$ and $T_2 = a T_1$,we substitute these values:$T_1 V_1^{\frac{7}{5}-1} = (a T_1) \left(\frac{V_1}{32}ight)^{\frac{7}{5}-1}$.$T_1 V_1^{2/5} = a T_1 \left(\frac{V_1}{32}ight)^{2/5}$.$1 = a \left(\frac{1}{32}ight)^{2/5}$.$1 = a \left(\frac{1}{2^5}ight)^{2/5}$.$1 = a \left(\frac{1}{2^2}ight) = a \left(\frac{1}{4}ight)$.Therefore,$a = 4$.

$A$ diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_1$ (in Kelvin) and the final temperature is $a T_1$,the value of $a$ is

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