The Celsius and Fahrenheit Scales — Question 3

Question 3

Doubling absolute temperature is not doubling Celsius

Use numerical readings $F=\frac95C+32$ and $T/\mathrm K=C+273.15$; these conversions are exact. A gas sample begins at $27.0^\circ\mathrm C$. Under a stipulated fixed-density ideal-gas model, pressure is proportional to absolute temperature.

1. Find the final Celsius and Fahrenheit readings required to double its pressure.

2. An operator instead doubles the Celsius reading to $54.0^\circ\mathrm C$. Find the actual pressure ratio and its percentage increase.

Question 3 - Solution

Doubling absolute temperature is not doubling Celsius

(a) Method and calculation

$$T_i=300.15\,\mathrm K,\qquad T_f=2T_i=600.30\,\mathrm K.$$$$C_f=327.15,\qquad F_f=1.8(327.15)+32=620.87.$$

(b) Interpretation and further calculation

$$\frac{p_f}{p_i}=\frac{54+273.15}{27+273.15}=\frac{327.15}{300.15}\approx1.08996.$$$$100\left(\frac{p_f}{p_i}-1\right)\approx8.996\%.$$A Celsius origin shift changes numerical ratios, so only the absolute-temperature ratio belongs in the supplied model.

Physical check

The correct doubled state is much hotter than $54^\circ$C; the large difference follows from the 273.15 K origin shift.