The volume-temperature graphs 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 $(m \propto \frac{1}{P})$,the line with the smallest slope corresponds to the highest pressure. Looking at the graph,the slope of the line for $p_1$ is the smallest,followed by $p_3$,and then $p_2$. Therefore,the correct order of pressures is $p_1 > p_3 > p_2$.

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

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