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(C) For an ideal gas at constant pressure,Charles's Law states that $V / T = 	ext{constant}.$ When the temperature changes from $T$ to $T + \Delta T,

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(C) For an ideal gas at constant pressure,Charles's Law states that $V / T = 	ext{constant}.$ When the temperature changes from $T$ to $T + \Delta T,$ the volume changes from $V$ to $V + \Delta V.$ Thus,we have: $\frac{V + \Delta V}{T + \Delta T} = \frac{V}{T}$ Cross-multiplying gives: $T(V + \Delta V) = V(T + \Delta T)$ $VT + T \Delta V = VT + V \Delta T$ Subtracting $VT$ from both sides: $T \Delta V = V \Delta T$ Rearranging to find the expression for $\delta$: $\delta = \frac{\Delta V}{V \Delta T} = \frac{1}{T}$ Since $\delta = 1/T,$ the quantity $\delta$ is inversely proportional to the temperature $T.$ This relationship is represented by a rectangular hyperbola,which corresponds to the graph shown in option $C.$

$A$. Process $A$	$P=5, V=7$
$B$. Process $B$	$P=4, V=6$
$C$. Process $C$	$P=12, V=1$
$D$. Process $D$	$P=6, V=3$

An ideal gas is initially at temperature $T$ and volume $V.$ Its volume is increased by $\Delta V$ due to an increase in temperature $\Delta T,$ pressure remaining constant. The quantity $\delta = \Delta V / (V \Delta T)$ varies with temperature as

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