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

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(B) For an adiabatic process,the relationship between temperature and volume is given by $TV^{\gamma-1} = 	ext{constant}$.Thus,$T_i V_i^{\gamma-1} =

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$4$

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(B) For an adiabatic process,the relationship between temperature and volume is given by $TV^{\gamma-1} = 	ext{constant}$.Thus,$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.Given $V_i = V_0$,$V_f = \frac{V_0}{32}$,and $\gamma = \frac{7}{5}$.Then $\gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5}$.Substituting the values: $T_i (V_0)^{2/5} = T_f \left(\frac{V_0}{32}ight)^{2/5}$.$T_f = T_i \left(\frac{V_0}{V_0/32}ight)^{2/5} = T_i (32)^{2/5}$.Since $32 = 2^5$,we have $T_f = T_i (2^5)^{2/5} = T_i (2^2) = 4T_i$.Comparing $T_f = xT_i$ with $T_f = 4T_i$,we get $x = 4$.

$A$ diatomic gas $\left(\gamma = \frac{7}{5}\right)$ is compressed adiabatically to volume $\frac{V_0}{32}$,where $V_0$ is its initial volume. The initial temperature of the gas is $T_i$ in Kelvin and the final temperature is $xT_i$ in Kelvin. The value of $x$ is:

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