Match each condition or graphical representation of a gas sample obeying Charles's law wit… — JAMB UTME

Match each condition or graphical representation of a gas sample obeying Charles's law with its corresponding physical or mathematical outcome.

A gas sample is heated from $27^\circ\text{C}$ to $54^\circ\text{C}$ at constant pressureVolume increases by a factor of $1.09$
A gas sample is heated from $300\text{ K}$ to $600\text{ K}$ at constant pressureVolume doubles ( $V_2 = 2V_1$ )
Extrapolation of a volume versus temperature ( $^\circ\text{C}$ ) plot to zero volumeIntersects the temperature axis at absolute zero ( $-273^\circ\text{C}$ )
Plot of volume ( $V$ ) versus absolute temperature ( $T$ in $\text{K}$ ) at constant pressureStraight line passing through the origin

Answer

The correct pairings match heating from 27 °C to 54 °C with a volume increase factor of 1.09, heating from 300 K to 600 K with volume doubling, extrapolation to zero volume with intersecting the temperature axis at absolute zero (-273 °C), and the V versus T (K) plot with a straight line through the origin.

Each item correctly links a concept of Charles's law to its outcome: heating from

27^\circ\text{C}

(

300\text{ K}

) to

54^\circ\text{C}

(

327\text{ K}

) increases volume by a factor of

1.09

; heating from

300\text{ K}

600\text{ K}

doubles absolute temperature and volume; extrapolating a

V

T

(

^\circ\text{C}

) graph to zero volume yields absolute zero (

-273^\circ\text{C}

); and plotting

V

against absolute temperature (

T

\text{K}

) produces a straight line passing through the origin.

Step-by-Step Solution

Convert Celsius temperatures to Kelvin for the first scenario

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 54 + 273 = 327\text{ K}

Charles's law requires temperatures to be expressed in Kelvin.

Determine the volume ratio for heating from

27^\circ\text{C}

54^\circ\text{C}

\frac{V_2}{V_1} = \frac{T_2}{T_1} = \frac{327\text{ K}}{300\text{ K}} = 1.09.

Doubling Celsius temperature does not double absolute temperature, so volume increases by a factor of 1.09.

Apply direct proportionality for the Kelvin heating scenario

Heating from

300\text{ K}

600\text{ K}

doubles the absolute temperature, so

V_2 = 2V_1

Volume is directly proportional to temperature on the Kelvin scale.

Analyze graphical characteristics of Charles's law

V

T

(K) plot is a straight line through the origin

(0,0)

, and extrapolating

V

T

(

^\circ\text{C}

) to zero volume yields absolute zero (

-273^\circ\text{C}

Absolute zero is the theoretical temperature at which gas volume extrapolates to zero.

Key Concept

Charles's Law and Absolute Temperature Scale

Estimated Time:1m 30s

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