A reversible adiabatic path on a P-V diagram | vedclass

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(C) For a reversible adiabatic process,the relation between pressure $P$ and volume $V$ is given by $P V^{\gamma} = 	ext{constant}$.Differentiating b

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$-2.0 	imes 10^7 \, N/m^5$

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(C) For a reversible adiabatic process,the relation between pressure $P$ and volume $V$ is given by $P V^{\gamma} = 	ext{constant}$.Differentiating both sides with respect to $V$,we get:$V^{\gamma} \frac{dP}{dV} + P \cdot \gamma V^{\gamma-1} = 0$Rearranging for the slope $\frac{dP}{dV}$:$\frac{dP}{dV} = -\frac{\gamma P}{V}$Given values are $P = 0.7 	imes 10^5 \, N/m^2$,$V = 0.0049 \, m^3$,and $\gamma = 1.4$.Substituting these values into the slope formula:$	ext{Slope} = -\frac{1.4 	imes (0.7 	imes 10^5)}{0.0049}$$	ext{Slope} = -\frac{0.98 	imes 10^5}{0.0049}$$	ext{Slope} = -\frac{98000}{0.0049} = -2.0 	imes 10^7 \, N/m^5$.

$A$ reversible adiabatic path on a $P-V$ diagram for an ideal gas passes through state $A$ where $P = 0.7 \times 10^5 \, N/m^2$ and $V = 0.0049 \, m^3$. The ratio of specific heats of the gas is $1.4$. The slope of the path at $A$ is:

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