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(C) For an adiabatic process, the relation between temperature $T$ and pressure $p$ is given by $T^\gamma p^{1-\gamma} = 	ext{constant}$.Taking the n

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$8.97 	imes 10^{-8} \ K$

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(C) For an adiabatic process, the relation between temperature $T$ and pressure $p$ is given by $T^\gamma p^{1-\gamma} = 	ext{constant}$.Taking the natural logarithm on both sides: $\gamma \ln T + (1-\gamma) \ln p = 	ext{constant}$.Differentiating both sides: $\gamma \frac{\Delta T}{T} + (1-\gamma) \frac{\Delta p}{p} = 0$.Rearranging for $\Delta T$: $\frac{\Delta T}{T} = \frac{\gamma - 1}{\gamma} \frac{\Delta p}{p}$.Given: $T = 273 \ K$ (at $NTP$), $\gamma = 1.5$, $\Delta p = 0.001 \ dyne/cm^2$, $p = 1.013 	imes 10^6 \ dyne/cm^2$.Substituting the values: $\Delta T = 273 	imes \left( \frac{1.5 - 1}{1.5} ight) 	imes \frac{0.001}{1.013 	imes 10^6}$.$\Delta T = 273 	imes \frac{0.5}{1.5} 	imes \frac{10^{-3}}{1.013 	imes 10^6} = 273 	imes \frac{1}{3} 	imes 0.987 	imes 10^{-9} \approx 8.97 	imes 10^{-8} \ K$.

$A$ sound wave passing through an ideal gas at $NTP$ produces a pressure change of $0.001 \ dyne/cm^2$ during adiabatic compression. The corresponding change in temperature $(\gamma = 1.5$ for the gas and atmospheric pressure is $1.013 \times 10^6 \ dyne/cm^2)$ is:

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