The expansion of unit mass of a perfect gas a | vedclass

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Question

(A) According to Charles's Law,for a fixed mass of a gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$

Vedclass · Accepted Answer

$a =$ volume,$b = ^{\circ}C$ temperature

Vedclass · Answer

(A) According to Charles's Law,for a fixed mass of a gas at constant pressure,the volume $V$ is directly proportional to the absolute temperature $T$ (in Kelvin),i.e.,$V = kT$.However,when expressed in Celsius scale,$V = V_0(1 + \alpha t)$,where $t$ is the temperature in $^{\circ}C$.This equation represents a straight line $y = mx + c$,where $y$ is volume $(V)$ and $x$ is temperature ($t$ in $^{\circ}C$).In the given graph,the line intersects the horizontal axis at a negative value,which corresponds to $-273.15^{\circ}C$ (absolute zero).Since volume cannot be negative,the vertical axis $(a)$ must represent Volume $(V)$ and the horizontal axis $(b)$ must represent temperature in $^{\circ}C$ because the temperature scale in $^{\circ}C$ can have negative values.Therefore,$a =$ volume and $b = ^{\circ}C$ temperature.

The expansion of unit mass of a perfect gas at constant pressure is shown in the diagram. Here

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