0
$\begingroup$

The coefficient of linear expansion is $$\Delta L = L \,\alpha \Delta T,$$ where $\Delta t$ can be the difference of any two temperatures. However, in volume expansion of gases, $$\Delta V = V \,\beta \Delta T,$$ and the difference has to be from 0 degree. Why?

$\endgroup$
1
  • $\begingroup$ I don't see why it has to be from 0 K. In both cases $\Delta T$ seems to refer to a temperature change. $\endgroup$ – user1704042 Nov 26 '14 at 3:46
1
$\begingroup$

Actually, that is not the correct formula for gases. In the usual model of ideal gases, the volume is given as a function of both temperature and pressure by the equation: $$V=\frac{n\bar RT}{p}$$

Both formulas you wrote are valid for solids and liquids (but not for gases). In those phases, volume is much more affected by temperature than by pressure, so the latter doesn't go into the equation.

The reason why the formulas you wrote (that, again, don't work for gases) only need temperature differences instead of absolute values is that for solids and liquids the temperature coefficients are often considered constant, so length or volume can be modeled in the form $\ell(T)=aT+b$ such that: $$\Delta\ell=\ell(T_2)-\ell(T_1)=(aT_2+b)-(aT_1+b)=a\Delta T$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.