Is thermal expansion a exponential function of temperature? Usually the thermal expansion is given as
$$V=V_{0}\alpha \Delta T$$
But the other solution, that is by solving the differential equation of the expansion I get an exponential function. But many source suggest that it is a linear function. Is it linear or exponential.
To be clear, the full form of equation of coefficient is
$$\alpha V = \frac{dV}{dT}$$
And thus solving it we get
$$V=V_{0}e^{\alpha T}$$
Am I wrong? And where I went wrong?
Edit : Sorry about that. V doesn't only signify volume, it signifies length and area too. Sorry for not mentioning it.
I am extremely sorry if I have committed any mistake in this. I am new to this one. Please mention that either.
 A: You have to allow for $\alpha_V$ to have temperature dependence. Take, as an example, the ideal gas. The equation of state implies that $V=Nk_BT/P$ expands linearly with $T$ at constant $P$. Indeed, the volume expansion coefficient is
$$
 \alpha_V = \frac{1}{V}\frac{dV}{dT}=\frac{1}{T},
$$
inversely proportional to $T$. This means $dV/V=dT/T$, and the expansion is linear.
A: Thomas has explained what happens for an ideal gas. Let's now consider a solid or liquid, for which your exponential solution is valid. Valid, yes, but it amounts to a linear solution because the expansivity of solids or liquids is so small that we can assume $\alpha T<<1$. In that case we can cut off the expansion of the exponential after the second term, so
$$ V=V_0e^{\alpha T} = V_0(1+\alpha T).$$
A: The equation looks correct! I think you may be reading sources that refer to linear expansion, which is expansion in one physical dimension (i.e., along a line $L$ instead of a volume $V$) but does not have $V$ as a linear function of $T$.
