A diatomic gas left(gamma = frac{7}{5}right) | vedclass

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

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$4$

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(B) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Given,initial volume $= V_i$,initial temperature $= T_i$,and final volume $V_f = \frac{V_i}{32}$.Using the adiabatic relation: $T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.Substituting the values: $T_i V_i^{\frac{7}{5}-1} = T_f \left(\frac{V_i}{32}ight)^{\frac{7}{5}-1}$.$T_i V_i^{\frac{2}{5}} = T_f \left(\frac{V_i}{32}ight)^{\frac{2}{5}}$.$T_i = T_f \left(\frac{1}{32}ight)^{\frac{2}{5}}$.$T_f = T_i 	imes (32)^{\frac{2}{5}}$.Since $32 = 2^5$,we have $(32)^{\frac{2}{5}} = (2^5)^{\frac{2}{5}} = 2^2 = 4$.Therefore,$T_f = 4T_i$.Comparing this with $T_f = xT_i$,we get $x = 4$.

$A$ diatomic gas $\left(\gamma = \frac{7}{5}\right)$ is compressed adiabatically to a volume $\frac{V_i}{32}$,where $V_i$ is its initial volume. The initial temperature of the gas is $T_i$ in Kelvin and the final temperature is $xT_i$. The value of $x$ is:

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