The graph of four gases with the same number | vedclass

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(A) From the ideal gas equation,$PV = \mu RT$,we can write:$P = \left( \frac{\mu R}{V} \right) T$Comparing this with the equation of a straight line $

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$V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$

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(A) From the ideal gas equation,$PV = \mu RT$,we can write:$P = \left( \frac{\mu R}{V} ight) T$Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = T$,the slope $m$ is given by:$m = 	an 	heta = \frac{\mu R}{V}$Since the number of moles $\mu$ and the gas constant $R$ are constant,we have:$V \propto \frac{1}{	an 	heta}$As the angle $	heta$ increases,$	an 	heta$ increases,and therefore the volume $V$ decreases.From the graph,the slope of line $1$ is equal to the slope of line $2$ (as they are parallel),so $V_1 = V_2$.Similarly,the slope of line $3$ is equal to the slope of line $4$,so $V_3 = V_4$.Since the slope of line $3$ (and $4$) is greater than the slope of line $1$ (and $2$),we have $	an 	heta_3 > 	an 	heta_1$.Therefore,$V_3  V_3$.Thus,the correct relation is $V_1 = V_2, V_3 = V_4$ and $V_2 > V_3$.

The graph of four gases with the same number of moles is given. Then:

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