The graph of pressure (P) versus density (d) | vedclass

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Question

(A) The ideal gas equation is $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$. Rearranging for pressure,$P = \frac{m}{V} \cdot \frac{

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$T_1 > T_2$

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(A) The ideal gas equation is $PV = nRT$. Since $n = \frac{m}{M}$,we have $PV = \frac{m}{M}RT$. Rearranging for pressure,$P = \frac{m}{V} \cdot \frac{RT}{M}$. Since density $d = \frac{m}{V}$,the equation becomes $P = d \cdot \frac{RT}{M}$. Comparing this with the equation of a straight line $y = mx$,where $y = P$ and $x = d$,the slope is $m = \frac{RT}{M}$. Since $R$ and $M$ are constants,the slope is directly proportional to temperature $T$. From the graph,the slope of the line for $T_1$ is greater than the slope of the line for $T_2$. Therefore,$T_1 > T_2$.

The graph of pressure $(P)$ versus density $(d)$ is shown for an ideal gas at two temperatures $T_1$ and $T_2$. Which of the following is correct?

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