The state of an ideal gas was changed isobari | vedclass

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(B) The equation of state for an ideal gas is $pV = nRT$,which can be rearranged as $V = (nR/p) \cdot T$.This represents the equation of a straight li

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$p_1 > p_2 > p_3$

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(B) The equation of state for an ideal gas is $pV = nRT$,which can be rearranged as $V = (nR/p) \cdot T$.This represents the equation of a straight line passing through the origin with a slope $m = nR/p$.From the relationship,we can see that the slope is inversely proportional to the pressure,i.e.,$	ext{slope} \propto 1/p$.By observing the given $V-T$ graph,the slope of the line for $p_3$ is the greatest,followed by $p_2$,and the slope for $p_1$ is the smallest.Since $	ext{slope} \propto 1/p$,a larger slope corresponds to a smaller pressure.Therefore,the relationship between the pressures is $p_1 > p_2 > p_3$.

The state of an ideal gas was changed isobarically. The graph depicts three such isobaric lines. Which of the following is true about the pressures of the gas?

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