JEE Chemistry States Of Matter MCQs Set 18

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Multiple Choice Questions

Question 1: A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin of the temperature scale. The plots at different values of pressure are shown. Which of the following orders of pressure is correct for this gas?
VolumeTemperaturep1p2p3p4 (all four lines start from the same point; \(p_1\) is the steepest and \(p_4\) the flattest)
(a) \(p_1\lt p_2=p_3\lt p_4\)
(b) \(p_1=p_2=p_3=p_4\)
(c) \(p_1>p_2>p_3>p_4\)
(d) \(p_1\lt p_2\lt p_3\lt p_4\)
Show Answer & Explanation

Answer: (d) \(p_1\lt p_2\lt p_3\lt p_4\)

Explanation:
1. At constant pressure, \(V=\frac{nR}{p}\,T\), so the slope of the V-T line is \(\frac{nR}{p}\).
2. A steeper line means a smaller pressure. \(p_1\) is steepest, so it has the lowest pressure.
3. Hence \(p_1\lt p_2\lt p_3\lt p_4\).

Teacher's Note:
1. Slope of V against T is nR over p: larger slope, smaller pressure.

Question 2: The given curve shows the variation of V/T as a function of T for a fixed mass and pressure P of an ideal gas (three horizontal lines are shown for pressures \(P_1\), \(P_2\), \(P_3\), with \(P_3\) the highest line and \(P_1\) the lowest). It follows from the curve that:
V/TTP3P2P1
(a) \(P_1>P_2>P_3\)
(b) \(P_1\lt P_2\lt P_3\)
(c) nothing can be predicted about the pressures
(d) \(P_1=P_2=P_3\)
Show Answer & Explanation

Answer: (a) \(P_1>P_2>P_3\)

Explanation:
1. For an ideal gas \(\frac VT=\frac{nR}{P}\), a constant for a fixed amount of gas and fixed pressure, so each curve is a horizontal line.
2. The higher the line, the smaller the pressure. \(P_3\) is the highest line, so it has the smallest pressure and \(P_1\) the largest.
3. Hence \(P_1>P_2>P_3\).

Teacher's Note:
1. V/T equals nR/P, so the highest line belongs to the lowest pressure.

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