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(A) The given equation for the thermodynamic process is $PV^{2.5} = 0.40$.To find the slope of the $P-V$ curve,we need to calculate the derivative $\f

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(A) The given equation for the thermodynamic process is $PV^{2.5} = 0.40$.To find the slope of the $P-V$ curve,we need to calculate the derivative $\frac{dP}{dV}$.First,express $P$ as a function of $V$:$P = 0.40 V^{-2.5}$Differentiating both sides with respect to $V$:$\frac{dP}{dV} = 0.40 	imes (-2.5) 	imes V^{-2.5-1}$$\frac{dP}{dV} = -1.0 	imes V^{-3.5}$Now,evaluate the slope at $V = 1 \, m^3$:$\left(\frac{dP}{dV}ight)_{V=1} = -1.0 	imes (1)^{-3.5}$$\left(\frac{dP}{dV}ight)_{V=1} = -1 \, Pa/m^3$Thus,the slope of the $P-V$ curve at $V = 1 \, m^3$ is $-1$.

An ideal gas is subjected to a thermodynamic process $PV^{2.5} = 0.40$,where $P$ is in $Pa$ and $V$ is in $m^3$. What is the slope of the $P-V$ curve with volume plotted against the $x-$axis at $V = 1 \, m^3$?

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