The log , T vs log , V curve is plotted for a | vedclass

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(C) For an adiabatic process, $TV^{\gamma-1} = 	ext{constant}$.Taking log on both sides: $\log \, T + (\gamma-1) \log \, V = 	ext{constant}$.This is

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Diatomic & undergoing an adiabatic change

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(C) For an adiabatic process, $TV^{\gamma-1} = 	ext{constant}$.Taking log on both sides: $\log \, T + (\gamma-1) \log \, V = 	ext{constant}$.This is the equation of a straight line $y = mx + c$, where $y = \log \, T$, $x = \log \, V$, and the slope $m = -(\gamma-1)$.From the graph, the slope $m = \frac{1.40 - 1.60}{1.60 - 1.10} = \frac{-0.20}{0.50} = -0.4$.Equating the slopes: $-(\gamma-1) = -0.4$, so $\gamma-1 = 0.4$, which gives $\gamma = 1.4$.For a diatomic gas, $\gamma = 1.4$ (e.g., $C_p/C_v = 7/5 = 1.4$).Thus, the gas is diatomic and undergoing an adiabatic change.

The $\log \, T$ vs $\log \, V$ curve is plotted for an ideal gas as shown. Which of the following is true for the gas?

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