Kinetic Particle Model of Matter: Question 10

Syllabus 2.1.3

Multiple choice Extended 1 mark

A fixed mass of gas is kept at a constant temperature. Its pressure, pp, is measured for several different volumes, VV, and a graph of pp (on the y-axis) against 1V\dfrac{1}{V} (on the x-axis) is plotted. What is the shape of this graph?

Choose an answer to check it, then compare with the worked solution below.

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Worked solution

Step 1: Start from the pressure–volume law

For a fixed mass of gas at constant temperature:

pV=constant,call it kpV = \text{constant}, \quad \text{call it } k

Step 2: Rearrange in terms of 1V\dfrac{1}{V}

Dividing both sides by VV:

p=k×1Vp = k \times \frac{1}{V}

Step 3: Recognise the form of this equation

This has the form p=k×xp = k \times x, where x=1Vx = \dfrac{1}{V}, a straight-line equation (y=mxy = mx) with gradient kk and no constant term added on. A graph of pp against 1V\dfrac{1}{V} is therefore a straight line whose gradient equals the constant kk (=pV=pV), and which passes through the origin, because when 1V=0\dfrac{1}{V} = 0, p=0p = 0.

Step 4: Match to the options

This matches option A.

Why the other options are wrong

OptionProblem
BThe equation p=k×(1/V)p = k \times (1/V) has no added constant, so the line must pass through the origin, not miss it.
CA curve approaching both axes is the shape of pp plotted directly against VV (not against 1/V1/V).
DA horizontal line would mean pressure never changes with volume, which contradicts pV=constantpV = \text{constant}.

Final answers

  • A graph of pp against 1V\dfrac{1}{V} for a fixed mass of gas at constant temperature is a straight line through the origin, option A.