From the following V-T diagram,we can conclud | vedclass

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(C) The given graph is a straight line not passing through the origin. The equation of the line is $V = mT + c$,where $m$ is the slope and $c$ is the

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$P_1 < P_2$

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(C) The given graph is a straight line not passing through the origin. The equation of the line is $V = mT + c$,where $m$ is the slope and $c$ is the intercept on the $V$-axis. Since the intercept $c$ is positive,we can write $V = aT + b$ (where $a, b > 0$).From the ideal gas equation,$PV = \mu RT$,we have $P = \frac{\mu RT}{V}$.Substituting $V = aT + b$,we get $P = \frac{\mu RT}{aT + b} = \frac{\mu R}{a + b/T}$.As $T$ increases,the term $b/T$ decreases,which means the denominator $(a + b/T)$ decreases.Since the denominator decreases as $T$ increases,the pressure $P$ must increase as $T$ increases.Given that $T_2 > T_1$,it follows that $P_2 > P_1$,or $P_1 < P_2$.

From the following $V-T$ diagram,we can conclude that:

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