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Exercises · 5.21

Q.In terms of Charles' law explain why –273 °C is the lowest possible temperature.

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Step 1 – State Charles' Law

At constant pressure and amount of gas:

V∝T(T in kelvin),orVt=V0(1+t273)(t in °C)V \propto T \quad (\text{T in kelvin}), \qquad \text{or} \qquad V_t = V_0\left(1+\frac{t}{273}\right) \quad (t \text{ in °C})

Step 2 – Examine the plot of VV vs. tt(°C)

Charles' law gives a straight line when volume is plotted against Celsius temperature. Real gas data, when extrapolated backward (below the point where the gas actually liquefies), always meets the temperature axis at the SAME point for every gas:

t=−273°C(i.e., V=0)t = -273°\text{C} \quad (\text{i.e., } V=0)

Step 3 – Physical reasoning: why this must be the lowest possible temperature

From Vt=V0(1+t/273)V_t = V_0(1+t/273): at t=−273°t=-273°C, Vt=V0(1−1)=0V_t = V_0(1-1) = 0.

At any temperature BELOW −273°-273°C, the formula would predict a NEGATIVE volume, which is physically impossible (a substance cannot occupy negative space). Since volume can never be less than zero, temperature can never go below −273°-273°C.

Step 4 – Conclusion …

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