The volume-temperature graph of a given mass | vedclass

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Question

(A) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (\frac{nR}{P})T$.This represents a straight line equation $y = mx$,

Vedclass · Accepted Answer

$P_1 > P_3 > P_2$

Vedclass · Answer

(A) According to the ideal gas equation,$PV = nRT$,which can be rearranged as $V = (\frac{nR}{P})T$.This represents a straight line equation $y = mx$,where the slope $m = \frac{nR}{P}$.Since the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$,a smaller slope corresponds to a higher pressure.From the given graph,the slope of the line for $P_1$ is the smallest,followed by $P_3$,and then $P_2$ (which has the largest slope).Therefore,the correct order of pressures is $P_1 > P_3 > P_2$.

Graph	Coordinates ($y$ vs $x$)
$(A)$ Hyperbola	$1. \ y = PV, x = V$
$(B)$ Straight line passing through origin	$2. \ y = P, x = V$
$(C)$ Horizontal line parallel to $x$-axis	$3. \ y = P, x = 1/V$

The volume-temperature graph of a given mass of an ideal gas at constant pressure is shown below. What is the correct order of pressures?

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